What is the best way to factor $25x^2-121$? [closed]
$begingroup$
What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
algebra-precalculus factoring
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
$endgroup$
closed as off-topic by abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd Dec 30 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 5 more comments
$begingroup$
What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
algebra-precalculus factoring
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
$endgroup$
closed as off-topic by abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd Dec 30 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Do you know about the phrase "difference of squares"? What method are you attempting to use?
$endgroup$
– abiessu
Dec 30 '18 at 2:50
3
$begingroup$
$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
$endgroup$
– Zober
Dec 30 '18 at 2:51
2
$begingroup$
"The method I have been t[aught] doesn't work for this problem" Which method is it?
$endgroup$
– Did
Dec 30 '18 at 2:52
1
$begingroup$
Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
$endgroup$
– fleablood
Dec 30 '18 at 3:01
1
$begingroup$
To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
$endgroup$
– timtfj
Dec 30 '18 at 3:10
|
show 5 more comments
$begingroup$
What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
algebra-precalculus factoring
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
$endgroup$
What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
algebra-precalculus factoring
algebra-precalculus factoring
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
edited Dec 30 '18 at 2:54
Eevee Trainer
8,39021439
8,39021439
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
asked Dec 30 '18 at 2:49
H. HogH. Hog
263
263
closed as off-topic by abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd Dec 30 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd Dec 30 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – abiessu, Did, Chinnapparaj R, Jack D'Aurizio, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Do you know about the phrase "difference of squares"? What method are you attempting to use?
$endgroup$
– abiessu
Dec 30 '18 at 2:50
3
$begingroup$
$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
$endgroup$
– Zober
Dec 30 '18 at 2:51
2
$begingroup$
"The method I have been t[aught] doesn't work for this problem" Which method is it?
$endgroup$
– Did
Dec 30 '18 at 2:52
1
$begingroup$
Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
$endgroup$
– fleablood
Dec 30 '18 at 3:01
1
$begingroup$
To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
$endgroup$
– timtfj
Dec 30 '18 at 3:10
|
show 5 more comments
2
$begingroup$
Do you know about the phrase "difference of squares"? What method are you attempting to use?
$endgroup$
– abiessu
Dec 30 '18 at 2:50
3
$begingroup$
$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
$endgroup$
– Zober
Dec 30 '18 at 2:51
2
$begingroup$
"The method I have been t[aught] doesn't work for this problem" Which method is it?
$endgroup$
– Did
Dec 30 '18 at 2:52
1
$begingroup$
Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
$endgroup$
– fleablood
Dec 30 '18 at 3:01
1
$begingroup$
To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
$endgroup$
– timtfj
Dec 30 '18 at 3:10
2
2
$begingroup$
Do you know about the phrase "difference of squares"? What method are you attempting to use?
$endgroup$
– abiessu
Dec 30 '18 at 2:50
$begingroup$
Do you know about the phrase "difference of squares"? What method are you attempting to use?
$endgroup$
– abiessu
Dec 30 '18 at 2:50
3
3
$begingroup$
$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
$endgroup$
– Zober
Dec 30 '18 at 2:51
$begingroup$
$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
$endgroup$
– Zober
Dec 30 '18 at 2:51
2
2
$begingroup$
"The method I have been t[aught] doesn't work for this problem" Which method is it?
$endgroup$
– Did
Dec 30 '18 at 2:52
$begingroup$
"The method I have been t[aught] doesn't work for this problem" Which method is it?
$endgroup$
– Did
Dec 30 '18 at 2:52
1
1
$begingroup$
Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
$endgroup$
– fleablood
Dec 30 '18 at 3:01
$begingroup$
Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
$endgroup$
– fleablood
Dec 30 '18 at 3:01
1
1
$begingroup$
To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
$endgroup$
– timtfj
Dec 30 '18 at 3:10
$begingroup$
To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
$endgroup$
– timtfj
Dec 30 '18 at 3:10
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.
Similar Question:
Let's factor the following: $$9x^4 - 16$$
Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;)
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)
Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":
- We begin with a trinomial, $ax^2 + bx + c$.
- We find factors of $a cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.
- Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.
- Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.
This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
$endgroup$
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
add a comment |
$begingroup$
The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $pm 1;mp 121$ or $pm 11, mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5xtimes 5x + 11times 5x + 5xtimes(-11) + 11times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
$endgroup$
1
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
1
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.
Similar Question:
Let's factor the following: $$9x^4 - 16$$
Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;)
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)
Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":
- We begin with a trinomial, $ax^2 + bx + c$.
- We find factors of $a cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.
- Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.
- Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.
This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
$endgroup$
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
add a comment |
$begingroup$
Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.
Similar Question:
Let's factor the following: $$9x^4 - 16$$
Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;)
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)
Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":
- We begin with a trinomial, $ax^2 + bx + c$.
- We find factors of $a cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.
- Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.
- Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.
This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
$endgroup$
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
add a comment |
$begingroup$
Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.
Similar Question:
Let's factor the following: $$9x^4 - 16$$
Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;)
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)
Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":
- We begin with a trinomial, $ax^2 + bx + c$.
- We find factors of $a cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.
- Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.
- Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.
This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
$endgroup$
Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.
Similar Question:
Let's factor the following: $$9x^4 - 16$$
Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;)
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)
Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":
- We begin with a trinomial, $ax^2 + bx + c$.
- We find factors of $a cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.
- Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.
- Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.
This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
edited Dec 30 '18 at 3:19
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
answered Dec 30 '18 at 2:52
Eevee TrainerEevee Trainer
8,39021439
8,39021439
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
add a comment |
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
$begingroup$
Special case of the AC method
$endgroup$
– Bill Dubuque
Dec 30 '18 at 20:06
add a comment |
$begingroup$
The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $pm 1;mp 121$ or $pm 11, mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5xtimes 5x + 11times 5x + 5xtimes(-11) + 11times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
$endgroup$
1
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
1
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
add a comment |
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The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $pm 1;mp 121$ or $pm 11, mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5xtimes 5x + 11times 5x + 5xtimes(-11) + 11times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
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1
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Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
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– John Omielan
Dec 30 '18 at 4:00
1
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That's what those emphasised IFs were for.
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– fleablood
Dec 30 '18 at 4:18
add a comment |
$begingroup$
The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $pm 1;mp 121$ or $pm 11, mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5xtimes 5x + 11times 5x + 5xtimes(-11) + 11times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
$endgroup$
The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $pm 1;mp 121$ or $pm 11, mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5xtimes 5x + 11times 5x + 5xtimes(-11) + 11times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
answered Dec 30 '18 at 3:18
fleabloodfleablood
73k22789
73k22789
1
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
1
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
add a comment |
1
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
1
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
1
1
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
$begingroup$
Note that in your general checks for how the values can be factored, you are assuming that the coefficients $a, b, c, d$ are integers, which doesn't have to be the case in general. However, no such restriction was provided in the problem statement (although it is implicit based on the use of perfect square coefficients), so you may wish to make it clear you are making this explicit assumption.
$endgroup$
– John Omielan
Dec 30 '18 at 4:00
1
1
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
$begingroup$
That's what those emphasised IFs were for.
$endgroup$
– fleablood
Dec 30 '18 at 4:18
add a comment |
2
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Do you know about the phrase "difference of squares"? What method are you attempting to use?
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– abiessu
Dec 30 '18 at 2:50
3
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$(5x)^2-(11)^2 = (5x-11)(5x+11)$.
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– Zober
Dec 30 '18 at 2:51
2
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"The method I have been t[aught] doesn't work for this problem" Which method is it?
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– Did
Dec 30 '18 at 2:52
1
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Hmm, trying to think of the best way to hint you toward discovering the difference of squares on your own.... Basically $a^2 - b^2 = (a+b)(a-b)$ and if we replaces $a^2$ with $25x^2=(5x)^2$ and $b^2 $ with $121=11^2$ we get $25x^2-121=(5x+11)(5x-11)$. But I don't want to tell you that... I want you to discover that on your own.... But I'm not sure the best hints to give you.
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– fleablood
Dec 30 '18 at 3:01
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To see where the "difference of two squares" formula comes from, just try multiplying out $(a+b)(a-b)$ and see what happens. It's incredibly useful.
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– timtfj
Dec 30 '18 at 3:10