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Q1W7 Learning Task 1 (Introduction)

Factoring Polynomials

On the chart below, find a factor in Column B of each of the given polynomials in Column A using the Factor Theorem.
Column A
1. x² + 6x + 8
2. x³ - 7x + 6
3. x³ - 2x² - 5x + 6
Column B
x-3
x-1
x+2
x+3
x+1

1 Answer

4 votes

Answer:

Column A Column B

1. x² + 6x + 8 x-3,x+2

2. x³ - 7x + 6 x+1, x+2, x+3

3. x³ - 2x² - 5x + 6 x-1, x+2, x-3

Explanation:

Column A Column B

1. x² + 6x + 8 x-3,x+2

2. x³ - 7x + 6 x+1, x+2, x+3

3. x³ - 2x² - 5x + 6 x-1, x+2, x-3

Using Factor theorem we put values of x = ±1,±2,±3 in each of the polynomials unless we get a zero.

1. x² + 6x + 8

= 1+6(1) +8= 15

1. x² + 6x + 8

4+ 12+8 = 24

1. x² + 6x + 8

(-1)² + 6(-1)+ 8

= 1-6+8= 3

1. x² + 6x + 8

(-2)² + 6(-2)+ 8

= 4-12+8= 0

1. x² + 6x + 8

(3)²+ 6(3) +8

= 9+18+8 ≠ 0

1. x² + 6x + 8

(-3)²+ 6(-3) +8

= 9-18+8 =-1

For this polynomial we have x+2= 0 or x=-2, x-3= 0 , x=3

2. x³ - 7x + 6

1-7+6= 0

2. x³ - 7x + 6

(-1)³-7(-1) +6

= 13-1≠0

2. x³ - 7x + 6

(2)³-7(2) +6

= 8-14+6= 0

2. x³ - 7x + 6

(-2)³-7(-2) +6

= -8 +14+6

2. x³ - 7x + 6

(-3)³-7(-3) +6

= -27+21+6 = 0

For this polynomial we have x+1= 0 , x+2 = 0 and x+3= 0, or x=-1,-2,-3

3. x³ - 2x² - 5x + 6

(1)³-2(1)²-5(1)+6

= 0

3. x³ - 2x² - 5x + 6

(-1)³-2(-1)²-5(-1)+6

= -1 -2 +5+6

=8

3. x³ - 2x² - 5x + 6

(2)³-2(2)²-5(2)+6

= 8-8-10+6

=-4

3. x³ - 2x² - 5x + 6

(-2)³-2(-2)²-5(-2)+6

= -8-8+10+6

=0

3. x³ - 2x² - 5x + 6

(3)³-2(3)²-5(3)+6

= 27-18-15+6

=0

3. x³ - 2x² - 5x + 6

(-3)³-2(-3)²-5(-3)+6

= -27-18+15+6

=-14

For this polynomial we have x-1= 0 ,x+2=0, x-3= 0or x=1,-2,3

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