# Unit 4- Factoring Polynomials

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How do you factor: 2x^2n + 7x^n – 15 ?

I'm confused because I've never worked it out when the equation has powers (^2n...) :/

Thanks!!!

I'm confused because I've never worked it out when the equation has powers (^2n...) :/

Thanks!!!

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#2

Hey Rosa

You know how you factorise 2x

You know how you factorise 2x

^{2}+ 7x - 15? You can literally do the same thing, and at the end just replace your xs with x^{n}s
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(Original post by

Hey Rosa

You know how you factorise 2x

**Student403**)Hey Rosa

You know how you factorise 2x

^{2}+ 7x - 15? You can literally do the same thing, and at the end just replace your xs with x^{n}sThank you!

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#4

I hope you understand the reason why btw!

Spoiler:

Show

Because 2x

You could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos

^{2n}= 2(x^{n})^{2}and so onYou could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos

^{2}θ +7cosθ - 15
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#5

(Original post by

How do you factor: 2x^2n + 7x^n – 15 ?

I'm confused because I've never worked it out when the equation has powers (^2n...) :/

Thanks!!!

**RosaA**)How do you factor: 2x^2n + 7x^n – 15 ?

I'm confused because I've never worked it out when the equation has powers (^2n...) :/

Thanks!!!

*disguised*quadratic (in this case, it's a quadratic in ). What this fancy shmoozy wording means is that if you make a substitution (I

*always*make this explicit substitution when working out problems like this) then you have a quadratic in or a quadratic in written as .

I'd make the substitution, re-write everything in terms of and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to by back-substitution .

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#7

**RosaA**)

How do you factor: 2x^2n + 7x^n – 15 ?

I'm confused because I've never worked it out when the equation has powers (^2n...) :/

Thanks!!!

Now, we can factorise this the regular way (i.e. the way we do quadratic equations as 403 mentioned)

Now, it seems that we have two equations from which we can obtain factors:

We can now evaluate these two equations to obtain our roots, so with the first one,

, as taking a square root gives us two solutions.

However, for the second equation, note that it leads to and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution .

We only have two solutions in this case:

I hope this example helped!

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#8

(Original post by

In addition to 403's excellent answer: This is called a

I'd make the substitution, re-write everything in terms of and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to by back-substitution .

**Zacken**)In addition to 403's excellent answer: This is called a

*disguised*quadratic (in this case, it's a quadratic in ). What this fancy shmoozy wording means is that if you make a substitution (I*always*make this explicit substitution when working out problems like this) then you have a quadratic in or a quadratic in written as .I'd make the substitution, re-write everything in terms of and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to by back-substitution .

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(Original post by

I'll show you an example - suppose the case where n = 2, and we have the equation .

Now, we can factorise this the regular way (i.e. the way we do quadratic equations as 403 mentioned)

Now, it seems that we have two equations from which we can obtain factors:

We can now evaluate these two equations to obtain our roots, so with the first one,

, as taking a square root gives us two solutions.

However, for the second equation, note that it leads to and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution .

We only have two solutions in this case:

I hope this example helped!

**aymanzayedmannan**)I'll show you an example - suppose the case where n = 2, and we have the equation .

Now, we can factorise this the regular way (i.e. the way we do quadratic equations as 403 mentioned)

Now, it seems that we have two equations from which we can obtain factors:

We can now evaluate these two equations to obtain our roots, so with the first one,

, as taking a square root gives us two solutions.

However, for the second equation, note that it leads to and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution .

We only have two solutions in this case:

I hope this example helped!

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**Zacken**)

In addition to 403's excellent answer: This is called a

*disguised*quadratic (in this case, it's a quadratic in ). What this fancy shmoozy wording means is that if you make a substitution (I

*always*make this explicit substitution when working out problems like this) then you have a quadratic in or a quadratic in written as .

I'd make the substitution, re-write everything in terms of and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to by back-substitution .

0

reply

(Original post by

My pleasure and thanks for the tag

I hope you understand the reason why btw!

**Student403**)My pleasure and thanks for the tag

I hope you understand the reason why btw!

Spoiler:

Show

Because 2x

You could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos

^{2n}= 2(x^{n})^{2}and so onYou could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos

^{2}θ +7cosθ - 15Yup, I understand it now

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