Find the roots of the polynomial: f(x)=x4+4x2−32

Question

lalilulelo2k3eq · Accepted Answer

Step 1
We have to find the roots of polynomial:

f (x) = x^{4} + 4 x^{2} - 32

To find the roots of given polynomial we need to put that polynomial equals to zero.
Therefore,
f(x)=0

x^{4} + 4 x^{2} - 32 = 0

Solving by factorization method,

x^{4} + 8 x^{2} - 4 x^{2} - 32 = 0

x^{2} (x^{2} + 8) - 4 (x^{2} + 8) = 0

(x^{2} + 8) (x^{2} - 4) = 0

Step 2
Either,

x^{2} + 8 = 0

x^{2} = - 8

x = \pm \sqrt{- 8}

= \pm \sqrt{8} i

(since

i = \sqrt{- 1}

, imaginary unit)

= \pm 2 \sqrt{2} i

= 2 \sqrt{2} i, - 2 \sqrt{2} i

or,

x^{2} - 4 = 0

x^{2} = 4

x = \pm \sqrt{4}

= \pm 2

=2, -2
Hence, roots of the polynomial are

x = - 2 \sqrt{2} i, - 2, 2, 2 \sqrt{2} i

.

Dawn Neal · Accepted Answer

Step 1  Explanation:  Given that,   f(x)=x4+4x2−32  Let us suppose   x2=u  Function becomes   u2+4u−32  Step 2  Now factorize the quadratic  u2+4u−32  (u+8)(u-4)  Recall the value of u  (x2+8)(x2−4)  Further, factorize the (x2−4)=(x−2)(x+2)  We get,  f(x)=(x2+8)(x−2)(x+2)

Find the roots of the polynomial: f(x)=x^{4}+4x^{2}-32

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