Question

Find all the zeros of the polynomial x⁴ + x³ – 34x² – 4x + 120, if two of its zeros are 2 and –2.

Answer 1

We know that if x = a is a zero of a polynomial, then x – a is a factor of f(x).

Since, 2 and –2 are zeros of f(x).

Therefore

(x + 2)(x – 2) = x² – 2²

= x² – 4

x² – 4 is a factor of f(x). Now, we divide x⁴ + x³ – 34x² – 4x + 120 by g(x) = x² – 4 to find the other zeros of f(x).

                    x² + x – 30
`x^2 - 4")"overline(+ \cancel(x^4) + x^3 - 34x^2 - 4x + 120)`
            `+\cancel(x^4) + 0 +     4x^2`
             –               –                          
                      `+ \cancel(x^3) - 30x^2 - \cancel(4x)`
                      `+ \cancel(x^3) +   0 - \cancel(4x)`
                       –                 +               
                                `- \cancel(30x^2) + \cancel(120)`
                                `- \cancel(30x^2) + \cancel(120)`
                                 +            –          
                                               0

By using that division algorithm we have,

f(x) = g(x) × q(x) – r(x)

x⁴ + x³ – 34x² – 4x + 120 = (x² – 4)(x² + x – 30) – 0

x⁴ + x³ – 34x² – 4x + 120 = (x + 2)(x – 2)(x² + 6x – 5x – 30)

x⁴ + x³ – 34x² – 4x + 120 = (x + 2)(x – 2)(x(x + 6) – 5(x + 6))

x⁴ + x³ – 34x² – 4x + 120 = (x + 2)(x – 2)(x + 6)(x – 5)

Hence, the zeros of the given polynomial are –2, +2, –6 and 5.