Question

apply

1. determine the value(s) of k so that the binomial is a factor of the polynomial.

a) x² - x + k, x - 2
b) x² - 6x - 7, x + k
c) x³ + 4x² + x + k, x + 2
d) x² + kx - 16, x - 2

Explanation:

Step1: Use the factor - theorem

If \(x - a\) is a factor of a polynomial \(P(x)\), then \(P(a)=0\).

a)

For the polynomial \(P(x)=x^{2}-x + k\) and the factor \(x - 2\), let \(x = 2\). Then \(P(2)=2^{2}-2 + k=0\).
\[

$$\begin{align*} 4-2 + k&=0\\ 2 + k&=0\\ k&=- 2 \end{align*}$$

\]

b)

For the polynomial \(P(x)=x^{2}-6x - 7\) and the factor \(x + k\) (i.e., \(x-(-k)\)), let \(x=-k\). Then \(P(-k)=(-k)^{2}-6(-k)-7 = 0\), which simplifies to \(k^{2}+6k - 7=0\).
Factor the quadratic equation: \((k + 7)(k - 1)=0\).
So \(k=-7\) or \(k = 1\).

c)

For the polynomial \(P(x)=x^{3}+4x^{2}+x + k\) and the factor \(x + 2\) (i.e., \(x-(-2)\)), let \(x=-2\). Then \(P(-2)=(-2)^{3}+4(-2)^{2}+(-2)+k=0\).
\[

$$\begin{align*} -8 + 16-2 + k&=0\\ 6 + k&=0\\ k&=-6 \end{align*}$$

\]

d)

For the polynomial \(P(x)=x^{2}+kx - 16\) and the factor \(x - 2\), let \(x = 2\). Then \(P(2)=2^{2}+2k-16=0\).
\[

$$\begin{align*} 4+2k-16&=0\\ 2k-12&=0\\ 2k&=12\\ k&=6 \end{align*}$$

\]

Answer:

a) \(k=-2\)
b) \(k=-7\) or \(k = 1\)
c) \(k=-6\)
d) \(k = 6\)