Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by+56$. To find $a$ and $b$, set up a system to be solved.
$$a+b=15$$ $$ab=1\times 56=56$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $56$.
$$1,56$$ $$2,28$$ $$4,14$$ $$7,8$$
Calculate the sum for each pair.
$$1+56=57$$ $$2+28=30$$ $$4+14=18$$ $$7+8=15$$
The solution is the pair that gives sum $15$.
$$a=7$$ $$b=8$$
Rewrite $y^{2}+15y+56$ as $\left(y^{2}+7y\right)+\left(8y+56\right)$.
$$\left(y^{2}+7y\right)+\left(8y+56\right)$$
Factor out $y$ in the first and $8$ in the second group.
$$y\left(y+7\right)+8\left(y+7\right)$$
Factor out common term $y+7$ by using distributive property.
$$\left(y+7\right)\left(y+8\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$y^{2}+15y+56=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$y=\frac{-15±\sqrt{15^{2}-4\times 56}}{2}$$
Square $15$.
$$y=\frac{-15±\sqrt{225-4\times 56}}{2}$$
Multiply $-4$ times $56$.
$$y=\frac{-15±\sqrt{225-224}}{2}$$
Add $225$ to $-224$.
$$y=\frac{-15±\sqrt{1}}{2}$$
Take the square root of $1$.
$$y=\frac{-15±1}{2}$$
Now solve the equation $y=\frac{-15±1}{2}$ when $±$ is plus. Add $-15$ to $1$.
$$y=-\frac{14}{2}$$
Divide $-14$ by $2$.
$$y=-7$$
Now solve the equation $y=\frac{-15±1}{2}$ when $±$ is minus. Subtract $1$ from $-15$.
$$y=-\frac{16}{2}$$
Divide $-16$ by $2$.
$$y=-8$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-7$ for $x_{1}$ and $-8$ for $x_{2}$.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$y^{2}+15y+56=\left(y+7\right)\left(y+8\right)$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.
$$x ^ 2 +15x +56 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = -15 $$ $$ rs = 56$$
Two numbers $r$ and $s$ sum up to $-15$ exactly when the average of the two numbers is $\frac{1}{2}*-15 = -\frac{15}{2}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
