Question

Factorize the following

$y^2+7y+12$

• A. $(y+3)(y+4)$
• B. $(y-3)(y+4)$
• C. $(y+3)(y-4)$
• D. None of these

Answer 1

Answer: (A)

$(y+3)(y+4)$

In the given polynomial $y^2+7y+12$,

The first term is $y^2$ and its coefficient is $1$.

The middle term is $7y$ and its coefficient is $7$.

The last term is a constant term $12$.

Multiply the coefficient of the first term by the constant $1\times 12=12$.

We now find the factor of $12$ whose sum equals the coefficient of the middle term, which is $7$ and then factorize the polynomial $y^2+7y+12$ as shown below:

$y^2+7y+12=y^2+4y+3y+12=y(y+4)+3(y+4)=(y+3)(y+4)$

Hence, $y^2+7y+12=(y+3)(y+4)$.