Question

which of the following is a factor of the cubic polynomial x³ - 10x² + 11x + 70? (1) x + 10 (2) x - 2 (3) x - 7 (4) x + 5

Explanation:

Step1: Apply the factor - theorem

According to the factor - theorem, if \(x - a\) is a factor of a polynomial \(P(x)\), then \(P(a)=0\). Let \(P(x)=x^{3}-10x^{2}+11x + 70\).

Step2: Test option (1) \(x + 10\) (i.e., \(a=-10\))

Substitute \(x=-10\) into \(P(x)\):
\[

$$\begin{align*} P(-10)&=(-10)^{3}-10\times(-10)^{2}+11\times(-10)+70\\ &=-1000 - 10\times100-110 + 70\\ &=-1000-1000-110 + 70\\ &=-2040 eq0 \end{align*}$$

\]

Step3: Test option (2) \(x - 2\) (i.e., \(a = 2\))

Substitute \(x = 2\) into \(P(x)\):
\[

$$\begin{align*} P(2)&=2^{3}-10\times2^{2}+11\times2+70\\ &=8-10\times4 + 22+70\\ &=8-40+22 + 70\\ &=60 eq0 \end{align*}$$

\]

Step4: Test option (3) \(x - 7\) (i.e., \(a = 7\))

Substitute \(x = 7\) into \(P(x)\):
\[

$$\begin{align*} P(7)&=7^{3}-10\times7^{2}+11\times7+70\\ &=343-10\times49+77 + 70\\ &=343-490+77+70\\ &=0 \end{align*}$$

\]

Step5: Test option (4) \(x + 5\) (i.e., \(a=-5\))

Substitute \(x=-5\) into \(P(x)\):
\[

$$\begin{align*} P(-5)&=(-5)^{3}-10\times(-5)^{2}+11\times(-5)+70\\ &=-125-10\times25-55 + 70\\ &=-125-250-55 + 70\\ &=-360 eq0 \end{align*}$$

\]

Answer:

(3) \(x - 7\)