Question

target c
one factor of (x^3 + 2x^2 - 23x - 60) is (x + 4). what are the remaining factors? show all work.
use powers in tools ((sum)) below to write in correct notation

Explanation:

Step1: Perform Polynomial Division

We divide the polynomial \(x^{3}+2x^{2}-23x - 60\) by the factor \(x + 4\) using polynomial long division.

First, divide the leading term of the dividend \(x^{3}\) by the leading term of the divisor \(x\), we get \(x^{2}\). Multiply the divisor \(x + 4\) by \(x^{2}\) to get \(x^{3}+4x^{2}\). Subtract this from the dividend:

\[

$$\begin{align*} &(x^{3}+2x^{2}-23x - 60)-(x^{3}+4x^{2})\\ =&x^{3}+2x^{2}-23x - 60 - x^{3}-4x^{2}\\ =&- 2x^{2}-23x - 60 \end{align*}$$

\]

Step2: Continue the Division

Now, divide the leading term of \(-2x^{2}\) by \(x\) to get \(-2x\). Multiply the divisor \(x + 4\) by \(-2x\) to get \(-2x^{2}-8x\). Subtract this from \(-2x^{2}-23x - 60\):

\[

$$\begin{align*} &(-2x^{2}-23x - 60)-(-2x^{2}-8x)\\ =&-2x^{2}-23x - 60 + 2x^{2}+8x\\ =&-15x - 60 \end{align*}$$

\]

Step3: Final Division Step

Divide the leading term of \(-15x\) by \(x\) to get \(-15\). Multiply the divisor \(x + 4\) by \(-15\) to get \(-15x-60\). Subtract this from \(-15x - 60\):

\[

$$\begin{align*} &(-15x - 60)-(-15x - 60)\\ =&-15x - 60 + 15x + 60\\ =&0 \end{align*}$$

\]

So, the quotient is \(x^{2}-2x - 15\).

Step4: Factor the Quotient

Now, we factor the quadratic \(x^{2}-2x - 15\). We need two numbers that multiply to \(-15\) and add up to \(-2\). The numbers are \(-5\) and \(3\).

So, \(x^{2}-2x - 15=(x - 5)(x + 3)\)

Answer:

The remaining factors are \(x - 5\) and \(x + 3\)