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)
+15001 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: \(x^2=u\\ x=\sqrt{u}\)
\(x_1=+\sqrt{16}\\ x_2=-\sqrt{16}\\ x_3=+\sqrt{1}\\ x_4=-\sqrt{1}\)
\(x_1=4\\ x_2=-4\\ x_3=1\\ x_4=-1\)
\(x^4-17x^2 +16=(x-4)(x+4)(x-1)(x+1)\)
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