Solve y^2-5y-14

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

Factor

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

Solution Steps

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-7 b=2$$

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

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

Factor out y in the first and 2 in the second group.

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

Factor out common term y-7 by using distributive property.

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

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.

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

Square -5.

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

Multiply -4 times -14.

$$y=\frac{-\left(-5\right)±\sqrt{25+56}}{2}$$

Add 25 to 56.

$$y=\frac{-\left(-5\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$y=\frac{-\left(-5\right)±9}{2}$$

The opposite of -5 is 5.

$$y=\frac{5±9}{2}$$

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

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

Divide 14 by 2.

$$y=7$$

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

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

Divide -4 by 2.

$$y=-2$$

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 -2 for x_{2}.

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

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

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

Evaluate

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