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Unit 4- Factoring Polynomials

RosaA

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

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

Thanks!!!

(edited 8 years ago)

Reply 1

Student403

Hey Rosa

You know how you factorise 2x² + 7x - 15? You can literally do the same thing, and at the end just replace your xs with xⁿs

Reply 2

RosaA

Original post by Student403

Hey Rosa

You know how you factorise 2x² + 7x - 15? You can literally do the same thing, and at the end just replace your xs with xⁿs

Okay, I've got it :smile:

Pretty simple, now that I understand :biggrin:

Thank you!

Reply 3

Student403

Original post by RosaA

Okay, I've got it :smile:

Pretty simple, now that I understand :biggrin:

Thank you!

My pleasure and thanks for the tag :biggrin:

I hope you understand the reason why btw!

Spoiler

(edited 8 years ago)

Reply 4

Zacken

Original post by RosaA

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

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

Thanks!!!

In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in

x^n

). 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)

u=x^n

then you have a quadratic in

u

or a quadratic in

x^n

written as

2u^2 + 7u - 15= 0

.

I'd make the substitution, re-write everything in terms of

u

and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to

x

by back-substitution

u=x^n

Reply 5

Zacken

Oh, and moved to maths.

Reply 6

Ayman!

Original post by RosaA

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

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

Thanks!!!

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

\displaystyle 2x^{4}+7x^{2}-15 = 0

.

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

\displaystyle 2x^{4}+ 10x^{2} - 3x^{2} -15= 0

\displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0

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

\displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0 \Longleftrightarrow 2x^{2}-3 = 0 \ \text{or} \ x^{2} +5 = 0

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

\displaystyle x^{2} = \frac{3}{2} \Longleftrightarrow x=\pm \sqrt{\left ( \frac{3}{2} \right )}

, as taking a square root gives us two solutions.

However, for the second equation, note that it leads to

\displaystyle x^{2} = -5

and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution

\displaystyle \Rightarrow x^{2} \neq -5

.

We only have two solutions in this case:

\displaystyle \therefore x=\pm \sqrt{\left ( \frac{3}{2} \right )}

I hope this example helped!

(edited 8 years ago)

Reply 7

Student403

Original post by Zacken

In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in

x^n

). 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)

u=x^n

then you have a quadratic in

u

or a quadratic in

x^n

written as

2u^2 + 7u - 15= 0

.

I'd make the substitution, re-write everything in terms of

u

and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to

x

by back-substitution

u=x^n

PRSOM

Reply 8

RosaA

Original post by aymanzayedmannan

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

\displaystyle 2x^{4}+7x^{2}-15 = 0

.

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

\displaystyle 2x^{4}+ 10x^{2} - 3x^{2} -15= 0

\displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0

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

\displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0 \Longleftrightarrow 2x^{2}-3 = 0 \ \text{or} \ x^{2} +5 = 0

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

\displaystyle x^{2} = \frac{3}{2} \Longleftrightarrow x=\pm \sqrt{\left ( \frac{3}{2} \right )}

, as taking a square root gives us two solutions.

However, for the second equation, note that it leads to

\displaystyle x^{2} = -5

and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution

\displaystyle \Rightarrow x^{2} \neq -5

.

We only have two solutions in this case:

\displaystyle \therefore x=\pm \sqrt{\left ( \frac{3}{2} \right )}

I hope this example helped!

Yes, it was very helpful indeed!! Thank you so much !! :biggrin:

Reply 9

RosaA

Original post by Zacken

In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in

x^n

). 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)