MHB Can you help me factor the expression x^4 - 15x^2 + 9?

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The expression x^4 - 15x^2 + 9 can be factored by substituting u = x^2, transforming it into u^2 - 15u + 9. A technique involving rewriting the expression as a difference of squares is discussed, leading to the observation that x^4 - 6x^2 + 9 can be factored as (x^2 - 3)^2 - (3x)^2. This results in the final factored form of (x^2 - 3 - 3x)(x^2 - 3 + 3x). The discussion highlights the learning experience and encourages further exploration of quadratic equations.
mathdad
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Factor the expression.

x^4 - 15x^2 + 9

Let u = x^2

Let x^4 = (x^2)^2

u^2 - 15u + 9

Must I use the quadratic formula here or completing the square?
 
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No, we want to use a trick kaliprasad showed us recently:

$$x^4-15x^2+9=x^4-6x^2+9-9x^2=\left(x^4-6x^2+9\right)-\left(9x^2\right)$$

Can you continue?
 
Can I apply the u-substitution?

Let u = x^2

(u^2 - 6u + 9) - 9u

(u - 3)(u - 3) - 9u

(x^2 - 3)(x^2 + 3) - 9x^2

Yes? No?
 
Last edited:
RTCNTC said:
Can I apply the u-substitution?

Let u = x^2

(u^2 - 6u + 9) - 9u

(u - 3)(u - 3) - 9u

(x^2 - 3)(x^2 + 3) - 9x^2

Yes? No?

What I intended for you to observe is that:

$$x^4-6x^2+9=\left(x^2-3\right)^2$$

$$9x^2=(3x)^2$$

And so the expression can be written as a difference of squares, and then factored as such. :)
 
I will be able to complete the factoring work here thanks to you. Keep in mind that I am now in the quadratic equations chapter of the David Cohen textbook. Lots of interesting questions in this chapter. I will be posting questions in terms of the discriminant, radical equations, literal equations and perhaps a few word problems.

I definitely know more math today since joining this website. Please, remind me to share with you what happened to me at Bank One in Springfield, MO 2006. The Bank One story is related to math and complete embarrassment. Look for a PM from me. I will text 5 general questions today or tomorrow.
 
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