Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
3. Dividing Polynomials
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Exercise 52 Page 682

What should the remainder of (x^2-x-k)Ă· (x+3) be if x+3 is a factor of x^2-x-k?

k=12

Practice makes perfect

We are asked to determine the value of k if x+3 is a factor of x^2-x- k. To do so, we will first divide x^2-x-k by x+3 and then examine the remainder of this division.

Division

We will use polynomial long division to perform the division. Both the dividend and the divisor are in standard form and the dividend has all terms present, so we do not need to rewrite the polynomials. Note that we will treat k as a number and not as a variable.
l r x + 3 & |l x^2 - x - k
â–Ľ
Divide

x^2/x= x

r x r x+3 & |l x^2 - x - k

Multiply term by divisor

r x rl x+3 & |l x^2 - x - k & x^2 +3x

Subtract down

r x r x+3 & |l - 4x-k
â–Ľ
Divide

- 4x/x= - 4

r x-4 r x+3 & |l - 4x-k

Multiply term by divisor

rx - 4 rl x+3 & |l - 4x-k & - 4x-12

Subtract down

r x-4 r x+3 & |l 12-k
We will stop the division here because the degree of the divisor is greater than the degree of the dividend. The quotient is x-4 with a remainder of 12-k.

Remainder

Using the result of the division, we can write the following equation. x^2-x-k/x+3=x-4+12-k/x+3 We can rewrite this equation such that the original polynomial is the sum of the divisor multiplied by the quotient and the remainder.
x^2-x-k/x+3=x-4+12-k/x+3
â–Ľ
LHS * ( x+3)=RHS* ( x+3)
x^2-x-k/x+3*(x+3)=(x-4+12-k/x+3)*(x+3)
x^2-x-k/x+3*(x+3)=(x-4)(x+3)+12-k/x+3*(x+3)
x^2-x-k=( x+3)(x-4)+( 12-k)
If x+3 is a factor of x^2-x-k, the remainder must be equal to 0. x^2-x-k=( x+3)(x-4) + 0 Therefore, we need to find the value of k which makes the expression 12-k equal to 0. 12-k=0 ⇔ k=12 If x+3 is a factor of x^2-x-k, the value of k must be 12.