Solve 5x^2+3x-8

$$5 { x }^{ 2 } +3x-8$$

Factor

$$\left(x-1\right)\left(5x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=3 ab=5\left(-8\right)=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite 5x^{2}+3x-8 as \left(5x^{2}-5x\right)+\left(8x-8\right).

$$\left(5x^{2}-5x\right)+\left(8x-8\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(x-1\right)+8\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(5x+8\right)$$

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.

$$5x^{2}+3x-8=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-8\right)}}{2\times 5}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 5\left(-8\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-3±\sqrt{9-20\left(-8\right)}}{2\times 5}$$

Multiply -20 times -8.

$$x=\frac{-3±\sqrt{9+160}}{2\times 5}$$

Add 9 to 160.

$$x=\frac{-3±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$x=\frac{-3±13}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-3±13}{10}$$

Now solve the equation x=\frac{-3±13}{10} when ± is plus. Add -3 to 13.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{-3±13}{10} when ± is minus. Subtract 13 from -3.

$$x=-\frac{16}{10}$$

Reduce the fraction \frac{-16}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -\frac{8}{5} for x_{2}.

$$5x^{2}+3x-8=5\left(x-1\right)\left(x-\left(-\frac{8}{5}\right)\right)$$

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

$$5x^{2}+3x-8=5\left(x-1\right)\left(x+\frac{8}{5}\right)$$

Add \frac{8}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+3x-8=5\left(x-1\right)\times \frac{5x+8}{5}$$

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

$$5x^{2}+3x-8=\left(x-1\right)\left(5x+8\right)$$

Evaluate

$$\left(x-1\right)\left(5x+8\right)$$