Question

question #4
which of the following polynomials has (x - 2) as a factor?
○ f(x)=x^4 - 3x^3 + x^2+3x - 2
○ f(x)=x^4 + 5x^3 + 9x^2 + 7x + 2
○ f(x)=x^4 + 3x^3 + x^2 - 3x - 2
○ f(x)=x^4 + x^3 - 3x^2 - x + 2
question #5
which of the following polynomials has (x + 1) as a factor?
○ f(x)=x^3 + 6x^2 + 3x - 10
○ f(x)=x^3 + 2x^2 - 13x + 10
○ f(x)=x^3 - 2x^2 - 13x - 10
○ f(x)=x^3 - 4x^2 - 7x + 10

Explanation:

Step1: Apply factor - theorem for Question #4

According to the factor - theorem, if \((x - a)\) is a factor of \(f(x)\), then \(f(a)=0\). For \((x - 2)\) as a factor, we substitute \(x = 2\) into each polynomial.

For \(f(x)=x^{4}-3x^{3}+x^{2}+3x - 2\):

\[

$$\begin{align*} f(2)&=2^{4}-3\times2^{3}+2^{2}+3\times2 - 2\\ &=16-24 + 4+6 - 2\\ &=0 \end{align*}$$

\]

For \(f(x)=x^{4}+5x^{3}+9x^{2}+7x + 2\):

\[

$$\begin{align*} f(2)&=2^{4}+5\times2^{3}+9\times2^{2}+7\times2 + 2\\ &=16+40+36 + 14+2\\ &=108 eq0 \end{align*}$$

\]

For \(f(x)=x^{4}+3x^{3}+x^{2}-3x - 2\):

\[

$$\begin{align*} f(2)&=2^{4}+3\times2^{3}+2^{2}-3\times2 - 2\\ &=16+24+4 - 6 - 2\\ &=36 eq0 \end{align*}$$

\]

For \(f(x)=x^{4}+x^{3}-3x^{2}-x + 2\):

\[

$$\begin{align*} f(2)&=2^{4}+2^{3}-3\times2^{2}-2 + 2\\ &=16+8-12 - 2+2\\ &=12 eq0 \end{align*}$$

\]
So the polynomial \(f(x)=x^{4}-3x^{3}+x^{2}+3x - 2\) has \((x - 2)\) as a factor.

Step2: Apply factor - theorem for Question #5

According to the factor - theorem, if \((x + 1)\) is a factor of \(f(x)\), then \(f(-1)=0\).

For \(f(x)=x^{3}+6x^{2}+3x - 10\):

\[

$$\begin{align*} f(-1)&=(-1)^{3}+6\times(-1)^{2}+3\times(-1)-10\\ &=-1 + 6-3-10\\ &=-8 eq0 \end{align*}$$

\]

For \(f(x)=x^{3}+2x^{2}-13x + 10\):

\[

$$\begin{align*} f(-1)&=(-1)^{3}+2\times(-1)^{2}-13\times(-1)+10\\ &=-1+2 + 13+10\\ &=24 eq0 \end{align*}$$

\]

For \(f(x)=x^{3}-2x^{2}-13x - 10\):

\[

$$\begin{align*} f(-1)&=(-1)^{3}-2\times(-1)^{2}-13\times(-1)-10\\ &=-1-2 + 13-10\\ &=0 \end{align*}$$

\]

For \(f(x)=x^{3}-4x^{2}-7x + 10\):

\[

$$\begin{align*} f(-1)&=(-1)^{3}-4\times(-1)^{2}-7\times(-1)+10\\ &=-1-4 + 7+10\\ &=12 eq0 \end{align*}$$

\]
So the polynomial \(f(x)=x^{3}-2x^{2}-13x - 10\) has \((x + 1)\) as a factor.

Answer:

Question #4: \(f(x)=x^{4}-3x^{3}+x^{2}+3x - 2\)
Question #5: \(f(x)=x^{3}-2x^{2}-13x - 10\)