Core Connections Integrated III, 2015
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Core Connections Integrated III, 2015 View details
1. Section 9.1
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Exercise 82 Page 457

Remember to use zeros as the coefficients for any degrees that are missing from the polynomial.

2x^4-2x+1+- 1/x-3

Practice makes perfect

To divide the given polynomials using synthetic division, we first need to rewrite the dividend so that all of the coefficients are present. Any missing terms should be added to the polynomial with a coefficient of 0. 2x^5-6x^4-2x^2+7x-4 ⇕ 2x^5-6x^4+ 0x^3-2x^2+7x-4 Remember that the general form of the synthetic division divisor must be x-a. Since the given divisor is already written in this form, we are ready to go.

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4

Bring down the first coefficient

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 &&&&& & c 2& & & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6& & & & & c 2& & & & &

Add down

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6& & & & & c 2& 0 & & & &
â–¼
Repeat the process for all of the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6&0 & & & & c 2& 0 & & & &

Add down

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6&0 & & & & c 2& 0 &0 & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6&0 & 0 & & & c 2& 0 & 0 & & &

Add down

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 &7 & -4 & 6&0 & 0 & & & c 2& 0 &0 &-2 & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 & 7 & -4 & 6&0 & 0 &- 6 & & c 2& 0 &0 & -2 & &

Add down

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 & 7 & -4 & 6&0 & 0 &- 6 & & c 2& 0 &0 &-2 & 1 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 & 7 & -4 & 6&0 & 0 &- 6 & 3 & c 2& 0 &0 &-2 & 1 &

Add down

rl IR-0.15cm r 3 & |rr 2 &-6 &0 &-2 & 7 & -4 & 6&0 & 0 &- 6 & 3 & c 2& 0 &0 &-2 & 1 & -1

The quotient is a polynomial of degree 4, with the above coefficients. The remainder is - 1. Quotient & Remainder 2x^4-2x+1 & - 1 We can see the above to rewrite the given expression. (2x^5-6x^4-2x^2+7x-4) ÷ (x-3) ⇕ 2x^4-2x+1+- 1/x-3

Checking Our Answer

Checking the answer
We can check our answer by multiplying the quotient by the divisor, and then adding the remainder. If the answer equals the dividend, it means our answer is correct. Let's do it!

( 2x^4-2x+1+- 1/x-3 ) (x-3)? =2x^5-6x^4-2x^2+7x-4
2x^4(x-3)-2x(x-3)+(x-3)+- (x-3)/x-3? =2x^5-6x^4-2x^2+7x-4
â–¼
Simplify left-hand side
2x^5-6x^4-2x(x-3)+(x-3)+- (x-3)/x-3? =2x^5-6x^4-2x^2+7x-4
2x^5-6x^4-2x^2+6x+(x-3)+- (x-3)/x-3? =2x^5-6x^4-2x^2+7x-4
2x^5-6x^4-2x^2+6x+(x-3)-x-3/x-3? =2x^5-6x^4-2x^2+7x-4
2x^5-6x^4-2x^2+6x+(x-3)-1? =2x^5-6x^4-2x^2+7x-4
x^5-6x^4-2x^2+7x-4= x^5-6x^4-2x^2+7x-4 ✓