Solve 7x^2-31x-20

$$7 { x }^{ 2 } -31x-20$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 7x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-31 ab=7\left(-20\right)=-140$$

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 -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.

$$a=-35 b=4$$

Rewrite 7x^{2}-31x-20 as \left(7x^{2}-35x\right)+\left(4x-20\right).

$$\left(7x^{2}-35x\right)+\left(4x-20\right)$$

Factor out 7x in the first and 4 in the second group.

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

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

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

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.

$$7x^{2}-31x-20=0$$

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.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 7\left(-20\right)}}{2\times 7}$$

Square -31.

$$x=\frac{-\left(-31\right)±\sqrt{961-4\times 7\left(-20\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-31\right)±\sqrt{961-28\left(-20\right)}}{2\times 7}$$

Multiply -28 times -20.

$$x=\frac{-\left(-31\right)±\sqrt{961+560}}{2\times 7}$$

Add 961 to 560.

$$x=\frac{-\left(-31\right)±\sqrt{1521}}{2\times 7}$$

Take the square root of 1521.

$$x=\frac{-\left(-31\right)±39}{2\times 7}$$

The opposite of -31 is 31.

$$x=\frac{31±39}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{31±39}{14}$$

Now solve the equation x=\frac{31±39}{14} when ± is plus. Add 31 to 39.

$$x=\frac{70}{14}$$

Divide 70 by 14.

$$x=5$$

Now solve the equation x=\frac{31±39}{14} when ± is minus. Subtract 39 from 31.

$$x=-\frac{8}{14}$$

Reduce the fraction \frac{-8}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{7}$$

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 -\frac{4}{7} for x_{2}.

$$7x^{2}-31x-20=7\left(x-5\right)\left(x-\left(-\frac{4}{7}\right)\right)$$

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

$$7x^{2}-31x-20=7\left(x-5\right)\left(x+\frac{4}{7}\right)$$

Add \frac{4}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$7x^{2}-31x-20=7\left(x-5\right)\times \frac{7x+4}{7}$$

Cancel out 7, the greatest common factor in 7 and 7.

$$7x^{2}-31x-20=\left(x-5\right)\left(7x+4\right)$$

Evaluate

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