Solve a^2+7a-60

$${ a }^{ 2 } +7a-60$$

Factor

$$\left(a-5\right)\left(a+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-60. To find p and q, set up a system to be solved.

$$p+q=7 pq=1\left(-60\right)=-60$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 7.

$$p=-5 q=12$$

Rewrite a^{2}+7a-60 as \left(a^{2}-5a\right)+\left(12a-60\right).

$$\left(a^{2}-5a\right)+\left(12a-60\right)$$

Factor out a in the first and 12 in the second group.

$$a\left(a-5\right)+12\left(a-5\right)$$

Factor out common term a-5 by using distributive property.

$$\left(a-5\right)\left(a+12\right)$$

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.

$$a^{2}+7a-60=0$$

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.

$$a=\frac{-7±\sqrt{7^{2}-4\left(-60\right)}}{2}$$

Square 7.

$$a=\frac{-7±\sqrt{49-4\left(-60\right)}}{2}$$

Multiply -4 times -60.

$$a=\frac{-7±\sqrt{49+240}}{2}$$

Add 49 to 240.

$$a=\frac{-7±\sqrt{289}}{2}$$

Take the square root of 289.

$$a=\frac{-7±17}{2}$$

Now solve the equation a=\frac{-7±17}{2} when ± is plus. Add -7 to 17.

$$a=\frac{10}{2}$$

Divide 10 by 2.

$$a=5$$

Now solve the equation a=\frac{-7±17}{2} when ± is minus. Subtract 17 from -7.

$$a=-\frac{24}{2}$$

Divide -24 by 2.

$$a=-12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -12 for x_{2}.

$$a^{2}+7a-60=\left(a-5\right)\left(a-\left(-12\right)\right)$$

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

$$a^{2}+7a-60=\left(a-5\right)\left(a+12\right)$$

Evaluate

$$\left(a-5\right)\left(a+12\right)$$