Question

rewrite (\frac{6x^4 - 7x^3 - x^2 + 35x}{3x^2 + 4x - 1}) in the form (q(x) + \frac{r(x)}{3x^2 + 4x - 1}), so (\frac{6x^4 - 7x^3 - x^2 + 35x}{3x^2 + 4x - 1} = square + \frac{square}{3x^2 + 4x - 1}). question help: video, read, written example. submit question

Explanation:

Step1: Divide leading terms

$\frac{6x^4}{3x^2}=2x^2$

Step2: Multiply divisor by $2x^2$

$2x^2(3x^2+4x-1)=6x^4+8x^3-2x^2$

Step3: Subtract from dividend

$(6x^4-7x^3-x^2+35x)-(6x^4+8x^3-2x^2)=-15x^3+x^2+35x$

Step4: Divide new leading terms

$\frac{-15x^3}{3x^2}=-5x$

Step5: Multiply divisor by $-5x$

$-5x(3x^2+4x-1)=-15x^3-20x^2+5x$

Step6: Subtract from new dividend

$(-15x^3+x^2+35x)-(-15x^3-20x^2+5x)=21x^2+30x$

Step7: Divide new leading terms

$\frac{21x^2}{3x^2}=7$

Step8: Multiply divisor by 7

$7(3x^2+4x-1)=21x^2+28x-7$

Step9: Subtract to get remainder

$(21x^2+30x)-(21x^2+28x-7)=2x+7$

Answer:

$\frac{6x^4 - 7x^3 - x^2 + 35x}{3x^2 + 4x - 1} = 2x^2 - 5x + 7 + \frac{2x + 7}{3x^2 + 4x - 1}$
First blank: $2x^2 - 5x + 7$
Second blank: $2x + 7$