Solve 5a^2-31a-28

$$5 { a }^{ 2 } -31a-28$$

Factor

$$\left(a-7\right)\left(5a+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5a^{2}+pa+qa-28. To find p and q, set up a system to be solved.

$$p+q=-31 pq=5\left(-28\right)=-140$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -31.

$$p=-35 q=4$$

Rewrite 5a^{2}-31a-28 as \left(5a^{2}-35a\right)+\left(4a-28\right).

$$\left(5a^{2}-35a\right)+\left(4a-28\right)$$

Factor out 5a in the first and 4 in the second group.

$$5a\left(a-7\right)+4\left(a-7\right)$$

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

$$\left(a-7\right)\left(5a+4\right)$$

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.

$$5a^{2}-31a-28=0$$

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.

$$a=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 5\left(-28\right)}}{2\times 5}$$

Square -31.

$$a=\frac{-\left(-31\right)±\sqrt{961-4\times 5\left(-28\right)}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-\left(-31\right)±\sqrt{961-20\left(-28\right)}}{2\times 5}$$

Multiply -20 times -28.

$$a=\frac{-\left(-31\right)±\sqrt{961+560}}{2\times 5}$$

Add 961 to 560.

$$a=\frac{-\left(-31\right)±\sqrt{1521}}{2\times 5}$$

Take the square root of 1521.

$$a=\frac{-\left(-31\right)±39}{2\times 5}$$

The opposite of -31 is 31.

$$a=\frac{31±39}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{31±39}{10}$$

Now solve the equation a=\frac{31±39}{10} when ± is plus. Add 31 to 39.

$$a=\frac{70}{10}$$

Divide 70 by 10.

$$a=7$$

Now solve the equation a=\frac{31±39}{10} when ± is minus. Subtract 39 from 31.

$$a=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$a=-\frac{4}{5}$$

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 -\frac{4}{5} for x_{2}.

$$5a^{2}-31a-28=5\left(a-7\right)\left(a-\left(-\frac{4}{5}\right)\right)$$

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

$$5a^{2}-31a-28=5\left(a-7\right)\left(a+\frac{4}{5}\right)$$

Add \frac{4}{5} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5a^{2}-31a-28=5\left(a-7\right)\times \frac{5a+4}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5a^{2}-31a-28=\left(a-7\right)\left(5a+4\right)$$

Evaluate

$$\left(a-7\right)\left(5a+4\right)$$