πŸ‘€

Answer :

Answer:

here you goes hope it helps you

Step-by-step explanation:

1

Common factor

βˆ’

3

2

+

7

+

2

0

-3y^{2}+7y+20

βˆ’3y2+7y+20

βˆ’

1

(

3

2

βˆ’

7

βˆ’

2

0

)

-1(3y^{2}-7y-20)

βˆ’1(3y2βˆ’7yβˆ’20)

2

Use the sum-product pattern

βˆ’

1

(

3

2

βˆ’

7

βˆ’

2

0

)

-1(3y^{2}{\color{#c92786}{-7y}}-20)

βˆ’1(3y2βˆ’7yβˆ’20)

βˆ’

1

(

3

2

+

5

βˆ’

1

2

βˆ’

2

0

)

-1(3y^{2}+{\color{#c92786}{5y}}{\color{#c92786}{-12y}}-20)

βˆ’1(3y2+5yβˆ’12yβˆ’20)

3

Common factor from the two pairs

βˆ’

1

(

3

2

+

5

βˆ’

1

2

βˆ’

2

0

)

-1(3y^{2}+5y-12y-20)

βˆ’1(3y2+5yβˆ’12yβˆ’20)

βˆ’

1

(

(

3

+

5

)

βˆ’

4

(

3

+

5

)

)

-1(y(3y+5)-4(3y+5))

βˆ’1(y(3y+5)βˆ’4(3y+5))

4

Rewrite in factored form

Solution

βˆ’

1

(

βˆ’

4

)

(

3

+

5

)

Go Teaching: Other Questions