Solve 8x^2-18x-35

$$8 { x }^{ 2 } -18x-35$$

Factor

$$\left(2x-7\right)\left(4x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-18 ab=8\left(-35\right)=-280$$

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 -280.

$$1,-280 2,-140 4,-70 5,-56 7,-40 8,-35 10,-28 14,-20$$

Calculate the sum for each pair.

$$1-280=-279 2-140=-138 4-70=-66 5-56=-51 7-40=-33 8-35=-27 10-28=-18 14-20=-6$$

The solution is the pair that gives sum -18.

$$a=-28 b=10$$

Rewrite 8x^{2}-18x-35 as \left(8x^{2}-28x\right)+\left(10x-35\right).

$$\left(8x^{2}-28x\right)+\left(10x-35\right)$$

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

$$4x\left(2x-7\right)+5\left(2x-7\right)$$

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

$$\left(2x-7\right)\left(4x+5\right)$$

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.

$$8x^{2}-18x-35=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 8\left(-35\right)}}{2\times 8}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 8\left(-35\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-18\right)±\sqrt{324-32\left(-35\right)}}{2\times 8}$$

Multiply -32 times -35.

$$x=\frac{-\left(-18\right)±\sqrt{324+1120}}{2\times 8}$$

Add 324 to 1120.

$$x=\frac{-\left(-18\right)±\sqrt{1444}}{2\times 8}$$

Take the square root of 1444.

$$x=\frac{-\left(-18\right)±38}{2\times 8}$$

The opposite of -18 is 18.

$$x=\frac{18±38}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{18±38}{16}$$

Now solve the equation x=\frac{18±38}{16} when ± is plus. Add 18 to 38.

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

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{18±38}{16} when ± is minus. Subtract 38 from 18.

$$x=-\frac{20}{16}$$

Reduce the fraction \frac{-20}{16} to lowest terms by extracting and canceling out 4.

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

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

$$8x^{2}-18x-35=8\left(x-\frac{7}{2}\right)\left(x-\left(-\frac{5}{4}\right)\right)$$

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

$$8x^{2}-18x-35=8\left(x-\frac{7}{2}\right)\left(x+\frac{5}{4}\right)$$

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

$$8x^{2}-18x-35=8\times \frac{2x-7}{2}\left(x+\frac{5}{4}\right)$$

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

$$8x^{2}-18x-35=8\times \frac{2x-7}{2}\times \frac{4x+5}{4}$$

Multiply \frac{2x-7}{2} times \frac{4x+5}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-18x-35=8\times \frac{\left(2x-7\right)\left(4x+5\right)}{2\times 4}$$

Multiply 2 times 4.

$$8x^{2}-18x-35=8\times \frac{\left(2x-7\right)\left(4x+5\right)}{8}$$

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

$$8x^{2}-18x-35=\left(2x-7\right)\left(4x+5\right)$$

Evaluate

$$\left(2x-7\right)\left(4x+5\right)$$