Polynomials Difficulty 4

TMUA practice: Polynomials, problem 1

A TMUA-calibre polynomials problem at difficulty 4 of 5, with the full worked solution.

The question

The roots of

x^3 + 2x^2 - x + 3

are

p, q,

and

r.

What is the value of

(p^2 + 4)(q^2 + 4)(r^2 + 4)?

64

75

100

125

144

The correct answer is highlighted. D

Worked solution

You can factor $(p^2 + 4)(q^2 + 4)(r^2 + 4)$ as $(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)$ .

For any polynomial $f(x)$ , you can create a new polynomial $f(x+2)$ , which will have roots that instead have the value subtracted.

Substituting $x-2$ and $x+2$ into $x$ for the first polynomial, gives you $10i-5$ and $-10i-5$ as $c$ for both equations. Multiplying $10i-5$ and $-10i-5$ together gives you $\boxed{\textbf{(D) }125}$ .

-ev2028