What is the quotient of $b^3 + 4b^2 – 3b + 126$ by b+7?

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Answer 1 · 2016-04-10T15:11:36

$b^2-3b+18$

Use long division, as used for integers, to find the quotient.

The divisor is $b+7$ .
Look at the first term of the dividend, i.e. $b^3$ .

What should be multiplied to $b$ (of the divisor) to get the first term of the dividend, i.e. $b^3$ ?

$bxx b^2=b^3$
Therefore, $b^2$ becomes the first term of the quotient.

Now, $b^2 xx (b+7)=b^3 + 7b^2$
Write it below the corresponding terms of the dividend and subtract.

We are now left with $-3b^2-3b+126$ .

Repeat.

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What is the quotient of b^3 + 4b^2 – 3b + 126b^3 + 4b^2 – 3b + 126 by b+7?