Question

Find all the roots of x^3-5x^2-7x+51 if one root is 4-i

Answer 1

All the roots are -3, 4-i and 4+i.

Explanation:

If oine root is 4 - i then another one is 4 + i as complex roots occur as conjugate pairs.

(4 - i)(4 + i)

= 16 - i^2

= 17.

As the last term = 51 = 3 * 17

looks like the other root is 3 or -3.

By the Factor theorem

If x = 3 then f(3) = 0

f(3) = 27 - 5(3)^2 - 7(3) + 51

= 27 - 45 - 21 + 51 = 12 so 3 is not a root.

If x = -3:

f(-3) = -27 - 45 _ 21 + 51

= 0

So, x = -3 is a root.

Answer 2

Hello,

Explanation:

P(x)=x^3-5x^2-7x+51

Since the coefficients are all reals,

4+i (conjugate of 4-i) is also a root.

The polynomial est divisible by (x-4-i)(x-4+i)=(x-4)²+1=x²-8x+17

If we divide P(x) by x²-8x+17 we find the quotient (x+3) and the remainder 0

P(x)=(x+4)(x-4-i(x-4+i)

Roots are -4,4+i and 4-i