Solve 3x^2+8x-16

$$3 { x }^{ 2 } +8x-16$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=3\left(-16\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite 3x^{2}+8x-16 as \left(3x^{2}-4x\right)+\left(12x-16\right).

$$\left(3x^{2}-4x\right)+\left(12x-16\right)$$

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

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

Factor out common term 3x-4 by using distributive property.

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

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.

$$3x^{2}+8x-16=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-16\right)}}{2\times 3}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3\left(-16\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12\left(-16\right)}}{2\times 3}$$

Multiply -12 times -16.

$$x=\frac{-8±\sqrt{64+192}}{2\times 3}$$

Add 64 to 192.

$$x=\frac{-8±\sqrt{256}}{2\times 3}$$

Take the square root of 256.

$$x=\frac{-8±16}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±16}{6}$$

Now solve the equation x=\frac{-8±16}{6} when ± is plus. Add -8 to 16.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{-8±16}{6} when ± is minus. Subtract 16 from -8.

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

Divide -24 by 6.

$$x=-4$$

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

$$3x^{2}+8x-16=3\left(x-\frac{4}{3}\right)\left(x-\left(-4\right)\right)$$

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

$$3x^{2}+8x-16=3\left(x-\frac{4}{3}\right)\left(x+4\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+8x-16=3\times \frac{3x-4}{3}\left(x+4\right)$$

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

$$3x^{2}+8x-16=\left(3x-4\right)\left(x+4\right)$$

Evaluate

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