2 years ago

Please helpppppp yallllll

1 answer:

8 0

Answer:6

Step-by-step explanation:

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nevsk [136]

Answer:

The answer is 4

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NeTakaya

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natima [27]

$\dfrac{x^3+10x^2+13x+39}{x^2+2x+1}$

$x^3=x\cdot x^2$ , and $x(x^2+2x+1)=x^3+2x^2+x$ . Subtracting this from the numerator gives a remainder of

$(x^3+10x^2+13x+39)-(x^3+2x^2+x)=8x^2+12x+39$

$8x^2=8\cdot x^2$ , and $8(x^2+2x+1)=8x^2+16x+8$ . Subtracting this from the previous remainder gives a new remainder of

$(8x^2+12x+39)-(8x^2+16x+8)=-4x+31$

$-84x$ is not a multiple of $x^2$ , so we're done. Then

$\dfrac{x^3+10x^2+13x+39}{x^2+2x+1}=x+8+\dfrac{-4x+31}{x^2+2x+1}$

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2 years ago