Question

3. Given the polynomial p(x) = x^4 - 2x^3 -7x^2 + 18x – 18 a. Without long division, find the remainder if P is divided by x+1. b. If one zero of P is 1-i, find the remaining zeros of P. c. Write P in factored form.

Answer 1

(a) remainder is -40

(b) The remaining zeroes are (x+3) and (x-3)

Explanation:

p(x) = x^4 - 2x^3 -7x^2 + 18x – 18

(a) Remainder of P(x) / (x+1) can be found using the remainder theorem, namely

let x + 1 = 0 => x = -1

remainder

= P(-1)

= (-1)^4 - 2(-1)^3 -7(-1)^2 + 18(-1) – 18

= 1 +2 -7-18-18

= -40

remainder is -40

(b)

If one zero is 1-i, then the conjugate 1+i is another zero.

in other words,

(x-1+i) and (x-1-i) are both factors.

whose product = (x^2-2x+2)

Divide p(x) by (x^2-2x+2) gives

p(x) by (x^2-2x+2)

= (x^4 - 2x^3 -7x^2 + 18x – 18) / (x^2-2x+2)

= x^2 -9

= (x+3) * (x-3)

The remaining zeroes are (x+3) and (x-3)