Question

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Factor completely, then place the factors in The proper location on the grid.3y2 +7y+4

Answer 1

We are asked to factor in the following expression:

$3y^2+7y+4$

To do that we will multiply by 3/3:

$3y^2+7y+4=\frac{3(3y^2+7y+4)}{3}$

Now, we use the distributive property on the numerator:

$\frac{3(3y^2+7y+4)}{3}=\frac{9y^2+7(3y)+12}{3}$

Now we factor in the numerator on the right side in the following form:

$\frac{9y^2+7(3y)+12}{3}=\frac{(3y+\cdot)(3y+\cdot)}{3}$

Now, in the spaces, we need to find 2 numbers whose product is 12 and their algebraic sum is 7. Those numbers are 4 and 3, since:

$\begin{gathered} 4\times3=12 \\ 4+3=7 \end{gathered}$

Substituting the numbers we get:

$\frac{(3y+4)(3y+3)}{3}$

Now we take 3 as a common factor on the parenthesis on the right:

$\frac{(3y+4)(3y+3)}{3}=\frac{(3y+4)3(y+1)}{3}$

Now we cancel out the 3:

$\frac{(3y+4)3(y+1)}{3}=(3y+4)(y+1)$

Therefore, the factored form of the expression is (3y + 4)(y + 1).