Math  /  Algebra

QuestionFactor the equation completely. y2+3y28y^{2}+3 y-28
Factored Form \square

Studdy Solution

STEP 1

1. We are given a quadratic expression y2+3y28 y^2 + 3y - 28 .
2. We need to factor this expression completely.
3. The expression is in the standard quadratic form ay2+by+c ay^2 + by + c .

STEP 2

1. Identify the coefficients a a , b b , and c c .
2. Use the factoring method for quadratic expressions.
3. Verify the factorization.

STEP 3

Identify the coefficients in the quadratic expression y2+3y28 y^2 + 3y - 28 .
Here, a=1 a = 1 , b=3 b = 3 , and c=28 c = -28 .

STEP 4

Find two numbers that multiply to a×c=1×(28)=28 a \times c = 1 \times (-28) = -28 and add to b=3 b = 3 .
The numbers are 7 7 and 4 -4 because 7×(4)=28 7 \times (-4) = -28 and 7+(4)=3 7 + (-4) = 3 .

STEP 5

Rewrite the middle term 3y 3y using the numbers found:
y2+7y4y28 y^2 + 7y - 4y - 28

STEP 6

Factor by grouping:
Group the terms: (y2+7y)+(4y28) (y^2 + 7y) + (-4y - 28) .
Factor each group:
y(y+7)4(y+7) y(y + 7) - 4(y + 7)

STEP 7

Factor out the common factor (y+7) (y + 7) :
(y4)(y+7) (y - 4)(y + 7)

STEP 8

Verify the factorization by expanding:
(y4)(y+7)=y2+7y4y28=y2+3y28 (y - 4)(y + 7) = y^2 + 7y - 4y - 28 = y^2 + 3y - 28
The factorization is verified as correct.
The factored form of the equation is:
(y4)(y+7) \boxed{(y - 4)(y + 7)}

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