Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
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Exercise 31 Page 349

Divide until the degree of the divisor is greater than the degree of the dividend.

x^2+6x+9

Practice makes perfect
To divide the given polynomials using polynomial long division, all the terms of the dividend must be present. Since there are no missing terms, we do not need to rewrite the polynomial. x^3+7x^2+15x+9Let's divide!
l r x + 1 & |l x^3 + 7x^2 + 15x + 9
Divide

x^3/x= x^2

r x^2 r x+1 & |l x^3 +7 x^2 + 15x + 9

Multiply term by divisor

r x^2 rl x+1 & |l x^3 +7 x^2 + 15x + 9 & x^3 +x^2

Subtract down

r x^2 r x+1 & |l 6x^2 +15 x + 9
Divide

6x^2/x= 6x

r x^2 +6x r x+1 & |l 6 x^2 + 15x + 9

Multiply term by divisor

rx^2 + 6x rl x+1 & |l 6 x^2 + 15x + 9 & 6x^2 + 6x

Subtract down

r x^2 +6x r x+1 & |l 9x + 9
Divide

9x/x= 9

r x^2 +6x +9 r x+1 & |l 9x + 9

Multiply term by divisor

rx^2 +6x + 9 rl x+1 & |l 9x +9 & 9x+9

Subtract down

r x^2 +6x +9 r x+1 & |l 0
The quotient is x^2+6x+9 with a remainder of 0.
We can check our answer. If it is correct, then the product of the quotient and the divisor, plus the remainder, will be equal to the dividend.

Showing Our Work

Long division by hand...

When we are doing long division by hand, it looks a bit different than how we have it in this solution. Here is how yours should look when you are writing it in your notebook.