Question

polynomial long division (level 3)
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use the long division method to find the result when (16x^3 + 24x^2 - 3x - 17) is divided by (4x + 3). if there is a remainder, express the result in the form (q(x) + \frac{r(x)}{b(x)}).
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Explanation:

Step1: Divide leading terms

$\frac{16x^3}{4x} = 4x^2$

Step2: Multiply divisor by $4x^2$

$4x^2(4x + 3) = 16x^3 + 12x^2$

Step3: Subtract from dividend

$(16x^3 + 24x^2 - 3x - 17) - (16x^3 + 12x^2) = 12x^2 - 3x - 17$

Step4: Divide new leading term

$\frac{12x^2}{4x} = 3x$? Correction: $\frac{12x^2}{4x}=3x$ → Wait, no: $12x^2 - 3x$ leading term $\frac{12x^2}{4x}=3x$? Wait, no, recheck: After first subtraction, remainder is $12x^2 -3x -17$. Next term: $\frac{12x^2}{4x}=3x$. Multiply divisor: $3x(4x+3)=12x^2+9x$. Subtract: $(12x^2-3x-17)-(12x^2+9x)=-12x-17$. Next term: $\frac{-12x}{4x}=-3$. Multiply divisor: $-3(4x+3)=-12x-9$. Subtract: $(-12x-17)-(-12x-9)=-8$. So quotient is $4x^2 +3x -3$? Wait no: First term $4x^2$, second term $3x$, third term $-3$. Yes. Remainder $-8$. So result is $4x^2 +3x -3 + \frac{-8}{4x+3}$? Wait wait initial first step: $16x^3/4x=4x^2$, correct. Multiply: $4x²*4x=16x³, 4x²*3=12x²$. Subtract: $24x²-12x²=12x²$, so next term is $12x²/4x=3x$, correct. Multiply $3x*4x=12x², 3x*3=9x$. Subtract: $-3x-9x=-12x$, correct. Next term: $-12x/4x=-3$, multiply: $-3*4x=-12x, -3*3=-9$. Subtract: $-17 - (-9)=-8$. Yes. So quotient is $4x² +3x -3$, remainder $-8$. So final form is $4x² +3x -3 + \frac{-8}{4x+3}$. Yes.

Wait correction on step4: Earlier typo, now fixed. So the correct answer is $4x^2 + 3x - 3 + \frac{-8}{4x + 3}$ which simplifies to $4x^2 + 3x - 3 - \frac{8}{4x + 3}$.

Step1: Divide leading terms

$\frac{16x^3}{4x} = 4x^2$

Step2: Multiply divisor by $4x^2$

$4x^2(4x + 3) = 16x^3 + 12x^2$

Step3: Subtract to get new remainder

$(16x^3 + 24x^2 - 3x -17) - (16x^3 +12x^2) =12x²-3x-17$

Step4: Next quotient term

$\frac{12x²}{4x}=3x$

Step5: Multiply divisor by $3x$

$3x(4x+3)=12x²+9x$

Step6: Subtract to get new remainder

$(12x²-3x-17)-(12x²+9x)=-12x-17$

Step7: Next quotient term

$\frac{-12x}{4x}=-3$

Step8: Multiply divisor by $-3$

$-3(4x+3)=-12x-9$

Step9: Subtract to get remainder

$(-12x-17)-(-12x-9)=-8$

Step10: Combine results

Quotient $4x²+3x-3$, remainder $-8$ → $4x²+3x-3 + \frac{-8}{4x+3}$ = $4x²+3x-3 - \frac{8}{4x+3}$

Answer:

$4x^2 - 3 + \frac{-8}{4x + 3}$