Solve y^2-14+5y

$$y ^ { 2 } - 14 + 5 y$$

Factor

$$\left(y-2\right)\left(y+7\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}+5y-14$$

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-14. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-14\right)=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite y^{2}+5y-14 as \left(y^{2}-2y\right)+\left(7y-14\right).

$$\left(y^{2}-2y\right)+\left(7y-14\right)$$

Factor out y in the first and 7 in the second group.

$$y\left(y-2\right)+7\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y+7\right)$$

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.

$$y^{2}+5y-14=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-14\right)}}{2}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$y=\frac{-5±\sqrt{25+56}}{2}$$

Add 25 to 56.

$$y=\frac{-5±\sqrt{81}}{2}$$

Take the square root of 81.

$$y=\frac{-5±9}{2}$$

Now solve the equation y=\frac{-5±9}{2} when ± is plus. Add -5 to 9.

$$y=\frac{4}{2}$$

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-5±9}{2} when ± is minus. Subtract 9 from -5.

$$y=-\frac{14}{2}$$

Divide -14 by 2.

$$y=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -7 for x_{2}.

$$y^{2}+5y-14=\left(y-2\right)\left(y-\left(-7\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$y^{2}+5y-14=\left(y-2\right)\left(y+7\right)$$

Evaluate

$$\left(y-2\right)\left(y+7\right)$$