Question

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Answer 1

Step-by-step explanation:

https://www.tiger-algebra.com/drill/(12y~2_28y_15)/(6y~2_35y_25)/(2y~2-y-3)/(2y~2_11y-20)/

STEP

1

:

Equation at the end of step 1

(((12•(y2))+28y)+15)

———————————————————— ÷ (((2•(y2))-y)-3) ÷ ((2y2+11y)-20)

(((6•(y2))+35y)+25)

STEP

2

:

Equation at the end of step

2

:

(((12•(y2))+28y)+15)

———————————————————— ÷ ((2y2-y)-3) ÷ (2y2+11y-20)

(((6•(y2))+35y)+25)

STEP

3

:

Equation at the end of step

3

:

(((12•(y2))+28y)+15)

———————————————————— ÷ (2y2-y-3) ÷ (2y2+11y-20)

(((2•3y2)+35y)+25)

STEP

4

:

Equation at the end of step

4

:

(((22•3y2)+28y)+15)

——————————————————— ÷ (2y2-y-3) ÷ (2y2+11y-20)

(6y2+35y+25)

STEP

5

:

12y2 + 28y + 15

Simplify ———————————————

6y2 + 35y + 25

Trying to factor by splitting the middle term

5.1 Factoring 12y2 + 28y + 15

The first term is, 12y2 its coefficient is 12 .

The middle term is, +28y its coefficient is 28 .

The last term, "the constant", is +15

Step-1 : Multiply the coefficient of the first term by the constant 12 • 15 = 180

Step-2 : Find two factors of 180 whose sum equals the coefficient of the middle term, which is 28 .