Solve 8x^2-20x-28

$$8 { x }^{ 2 } -20x-28$$

Factor

$$4\left(2x-7\right)\left(x+1\right)$$

Solution Steps

Factor out 4.

$$4\left(2x^{2}-5x-7\right)$$

Consider 2x^{2}-5x-7. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\left(-7\right)=-14$$

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 2x^{2}-5x-7 as \left(2x^{2}-7x\right)+\left(2x-7\right).

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

Factor out x in 2x^{2}-7x.

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

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$4\left(2x-7\right)\left(x+1\right)$$

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.

$$8x^{2}-20x-28=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 8\left(-28\right)}}{2\times 8}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 8\left(-28\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-20\right)±\sqrt{400-32\left(-28\right)}}{2\times 8}$$

Multiply -32 times -28.

$$x=\frac{-\left(-20\right)±\sqrt{400+896}}{2\times 8}$$

Add 400 to 896.

$$x=\frac{-\left(-20\right)±\sqrt{1296}}{2\times 8}$$

Take the square root of 1296.

$$x=\frac{-\left(-20\right)±36}{2\times 8}$$

The opposite of -20 is 20.

$$x=\frac{20±36}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{20±36}{16}$$

Now solve the equation x=\frac{20±36}{16} when ± is plus. Add 20 to 36.

$$x=\frac{56}{16}$$

Reduce the fraction \frac{56}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{20±36}{16} when ± is minus. Subtract 36 from 20.

$$x=-\frac{16}{16}$$

Divide -16 by 16.

$$x=-1$$

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

$$8x^{2}-20x-28=8\left(x-\frac{7}{2}\right)\left(x-\left(-1\right)\right)$$

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

$$8x^{2}-20x-28=8\left(x-\frac{7}{2}\right)\left(x+1\right)$$

Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-20x-28=8\times \frac{2x-7}{2}\left(x+1\right)$$

Cancel out 2, the greatest common factor in 8 and 2.

$$8x^{2}-20x-28=4\left(2x-7\right)\left(x+1\right)$$

Evaluate

$$4\left(2x-7\right)\left(x+1\right)$$