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Sagot :
We are asked to factor in the following expression:
[tex]3y^2+7y+4[/tex]To do that we will multiply by 3/3:
[tex]3y^2+7y+4=\frac{3(3y^2+7y+4)}{3}[/tex]Now, we use the distributive property on the numerator:
[tex]\frac{3(3y^2+7y+4)}{3}=\frac{9y^2+7(3y)+12}{3}[/tex]Now we factor in the numerator on the right side in the following form:
[tex]\frac{9y^2+7(3y)+12}{3}=\frac{(3y+\cdot)(3y+\cdot)}{3}[/tex]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:
[tex]\begin{gathered} 4\times3=12 \\ 4+3=7 \end{gathered}[/tex]Substituting the numbers we get:
[tex]\frac{(3y+4)(3y+3)}{3}[/tex]Now we take 3 as a common factor on the parenthesis on the right:
[tex]\frac{(3y+4)(3y+3)}{3}=\frac{(3y+4)3(y+1)}{3}[/tex]Now we cancel out the 3:
[tex]\frac{(3y+4)3(y+1)}{3}=(3y+4)(y+1)[/tex]Therefore, the factored form of the expression is (3y + 4)(y + 1).
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