Question

Match each polynomial with its factorization. 1. y 2 - y - 2 (4y - 3)(y + 2) 2. 3y 2 - 27 3(y - 3)(y + 3) 3. y 4 - 16 -4(y + 2)( y - 4) 4. y 2 - 12y + 36 (y + 1)(y - 2) 5. -4y 2 + 8y + 32 (y - 2)(y + 2)(y2 + 4) 6. 5y - 2 + 4y 2 - 4 (y - 6)2

Answer 1

Answer:

1. y^2 - y - 2 = (y + 1)(y - 2)

2. 3y^2 - 27 = 3(y - 3)(y + 3)

3. y^4 - 16 = (y - 2)(y + 2)(y^2 + 4)

4. y^2 - 12y + 36 = (y - 6)^2

5. -4y^2 + 8y + 32 = -4(y + 2)( y - 4)

6. 5y - 2 + 4y^2 - 4 = (4y - 3)(y + 2)          

Step-by-step explanation:

1. y^2 - y - 2=(y - 2)(y + 1)→

y^2 - y - 2 = (y + 1)(y - 2)

2. 3y^2 - 27

Common factor 3:

$3y^2-27=3(y^2-9)$

Using Difference of squares:

$a^2-b^2=(a-b)(a+b)$

with:

$a^2=y^2\\ \sqrt{a^2}=\sqrt{y^2}\\ a=y$

$b^2=9\\ \sqrt{b^2}=\sqrt{9}\\ b=3$

3y^2 - 27=3(y - 3)(y + 3)

3. y^4 - 16

Using Difference of squares:

$a^2-b^2=(a-b)(a+b)$

with:

$a^2=y^4\\ \sqrt{a^2}=\sqrt{y^4}\\ a=y^2$

$b^2=16\\ \sqrt{b^2}=\sqrt{16}\\ b=4$

y^4 - 16 = (y^2-4)(y^2+4)

Using Difference of squares in the first parentheses:

$a^2-b^2=(a-b)(a+b)$

with:

$a^2=y^2\\ \sqrt{a^2}=\sqrt{y^2}\\ a=y$

$b^2=4\\ \sqrt{b^2}=\sqrt{4}\\ b=2$

y^4 - 16 = (y - 2)(y + 2)(y^2 + 4)

4. y^2 - 12y + 36 = (y - 6)(y - 6)

y^2 - 12y + 36 = (y - 6)^2

5. -4y^2 + 8y + 32

Common factor -4:

$-4y^2+8y+32=-4(y^2-2y-8)\\ -4y^2+8y+32=-4(y-4)(y+2)$

-4y^2 - 12y + 32 = -4(y + 2)( y - 4)

6. 5y - 2 + 4y^2 - 4

Adding like terms:

5y - 2 + 4y^2 - 4 = 5y + 4y^2 - 6

Ordering the terms:

5y - 2 + 4y^2 - 4 = 4y^2 + 5y - 6

Writing 5y like: 8y - 3y = -3y + 8y

5y - 2 + 4y^2 - 4 = 4y^2 - 3y + 8y - 6

Grouping terms:

5y - 2 + 4y^2 - 4 = (4y^2 - 3y) + (8y - 6)

Common factor in the first parentheses y and 2 in the second parentheses:

5y - 2 + 4y^2 - 4 = y(4y - 3) + 2(4y - 3)

Common factor 4y-3:

5y - 2 + 4y^2 - 4 = (4y - 3)(y + 2)