QUESTION IMAGE
Question
sing values below to find the quotient when $4x^3 + x^2 - x + 2$. if there is a remainder, express the result in the for note: you may not need to use all the boxes. \
pt 1 out of 2 \
$- 23x - 24) \div (x + 2) = 4x^2 - 7x - 9 + -\frac{6}{x + 2}$
Step1: Perform Polynomial Long Division
We divide the polynomial \(4x^3 + x^2 - 23x - 24\) by \(x + 2\).
- First term: \(\frac{4x^3}{x}=4x^2\). Multiply \(x + 2\) by \(4x^2\) to get \(4x^3+8x^2\). Subtract this from the original polynomial: \((4x^3 + x^2 - 23x - 24)-(4x^3+8x^2)=-7x^2 - 23x - 24\).
Step2: Next Term
Now, divide \(-7x^2\) by \(x\) to get \(-7x\). Multiply \(x + 2\) by \(-7x\) to get \(-7x^2-14x\). Subtract this from \(-7x^2 - 23x - 24\): \((-7x^2 - 23x - 24)-(-7x^2-14x)=-9x - 24\).
Step3: Next Term
Divide \(-9x\) by \(x\) to get \(-9\). Multiply \(x + 2\) by \(-9\) to get \(-9x-18\). Subtract this from \(-9x - 24\): \((-9x - 24)-(-9x-18)=-6\).
Step4: Express the Result
The division gives us the quotient \(4x^2 - 7x - 9\) and a remainder of \(-6\). So, we express the result as \(4x^2 - 7x - 9+\frac{-6}{x + 2}\).
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\(4x^2 - 7x - 9-\frac{6}{x + 2}\) (or \(4x^2 - 7x - 9+\frac{-6}{x + 2}\))