Solve 4x+6x^2-16

$$4 x + 6 x ^ { 2 } - 16$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 2x+3x^{2}-8. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=2 ab=3\left(-8\right)=-24$$

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 -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 2.

$$a=-4 b=6$$

Rewrite 3x^{2}+2x-8 as \left(3x^{2}-4x\right)+\left(6x-8\right).

$$\left(3x^{2}-4x\right)+\left(6x-8\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}+4x-16=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6\left(-16\right)}}{2\times 6}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 6\left(-16\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-4±\sqrt{16-24\left(-16\right)}}{2\times 6}$$

Multiply -24 times -16.

$$x=\frac{-4±\sqrt{16+384}}{2\times 6}$$

Add 16 to 384.

$$x=\frac{-4±\sqrt{400}}{2\times 6}$$

Take the square root of 400.

$$x=\frac{-4±20}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-4±20}{12}$$

Now solve the equation x=\frac{-4±20}{12} when ± is plus. Add -4 to 20.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-4±20}{12} when ± is minus. Subtract 20 from -4.

$$x=-\frac{24}{12}$$

Divide -24 by 12.

$$x=-2$$

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 -2 for x_{2}.

$$6x^{2}+4x-16=6\left(x-\frac{4}{3}\right)\left(x-\left(-2\right)\right)$$

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

$$6x^{2}+4x-16=6\left(x-\frac{4}{3}\right)\left(x+2\right)$$

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

$$6x^{2}+4x-16=6\times \frac{3x-4}{3}\left(x+2\right)$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}+4x-16=2\left(3x-4\right)\left(x+2\right)$$

Evaluate

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