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# factoring question

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Show all work to factor x^4 − 17x^2 + 16 completely.

Aug 12, 2020

#1
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Factor the following:
x^4-17 x^2+16

x^4-17 x^2+16==(x^2)^2-17 x^2+16:
(x^2)^2-17 x^2+16

The factors of 16 that sum to -17 are -1 and -16. So, (x^2)^2-17 x^2+16==(x^2-1) (x^2-16):
(x^2-1) (x^2-16)

x^2-16==x^2-4^2:
(x^2-1) (x^2-4^2)

Factor the difference of two squares. x^2-4^2==(x-4) (x+4):
(x-4) (x+4) (x^2-1)

x^2-1==x^2-1^2:
(x^2-1^2) (x-4) (x+4)

Factor the difference of two squares. x^2-1^2==(x-1) (x+1):

(x-1) (x+1) (x-4) (x+4)

Aug 12, 2020
#2
+12696
+1

Show all work to factor x^4 − 17x^2 + 16 completely.

Hello Guest!

$$x^4-17x^2 +16=0$$

We determine the zeros.

Exchange:  $$x^2=u$$

$$u^2-17u+16=0\\ u^2-m\cdot u+n =0$$

$$u=-\frac{m}{2}\pm \sqrt{(\frac{m}{2})^2-n}\\ u=8.5\pm \sqrt{8.5^2-16}\\ u=8.5\pm7.5$$

$$u_1=16\\ u_2=1$$

Reinstate:  $$x^2=u\\ x=\sqrt{u}$$

$$x_1=+\sqrt{16}\\ x_2=-\sqrt{16}\\ x_3=+\sqrt{1}\\ x_4=-\sqrt{1}$$

$$x_1=4\\ x_2=-4\\ x_3=1\\ x_4=-1$$

$$x^4-17x^2 +16=(x-4)(x+4)(x-1)(x+1)$$

!

Aug 12, 2020
edited by asinus  Aug 12, 2020