Solve 3x^2-10x-8

$$3 { x }^{ 2 } -10x-8$$

Factor

$$\left(x-4\right)\left(3x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\left(-8\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=-12 b=2$$

Rewrite 3x^{2}-10x-8 as \left(3x^{2}-12x\right)+\left(2x-8\right).

$$\left(3x^{2}-12x\right)+\left(2x-8\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(x-4\right)+2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(3x+2\right)$$

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.

$$3x^{2}-10x-8=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 3}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±14}{6}$$

Now solve the equation x=\frac{10±14}{6} when ± is plus. Add 10 to 14.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{10±14}{6} when ± is minus. Subtract 14 from 10.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -\frac{2}{3} for x_{2}.

$$3x^{2}-10x-8=3\left(x-4\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$3x^{2}-10x-8=3\left(x-4\right)\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-10x-8=3\left(x-4\right)\times \frac{3x+2}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-10x-8=\left(x-4\right)\left(3x+2\right)$$

Evaluate

$$\left(x-4\right)\left(3x+2\right)$$