Solve 4x^2+16x+15

$$4 { x }^{ 2 } +16x+15$$

Factor

$$\left(2x+3\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=16 ab=4\times 15=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=6 b=10$$

Rewrite 4x^{2}+16x+15 as \left(4x^{2}+6x\right)+\left(10x+15\right).

$$\left(4x^{2}+6x\right)+\left(10x+15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(2x+3\right)+5\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(2x+5\right)$$

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.

$$4x^{2}+16x+15=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\times 15}}{2\times 4}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-16±\sqrt{256-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$x=\frac{-16±\sqrt{256-240}}{2\times 4}$$

Add 256 to -240.

$$x=\frac{-16±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$x=\frac{-16±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-16±4}{8}$$

Now solve the equation x=\frac{-16±4}{8} when ± is plus. Add -16 to 4.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-16±4}{8} when ± is minus. Subtract 4 from -16.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{2} for x_{1} and -\frac{5}{2} for x_{2}.

$$4x^{2}+16x+15=4\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$4x^{2}+16x+15=4\left(x+\frac{3}{2}\right)\left(x+\frac{5}{2}\right)$$

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

$$4x^{2}+16x+15=4\times \frac{2x+3}{2}\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+16x+15=4\times \frac{2x+3}{2}\times \frac{2x+5}{2}$$

Multiply \frac{2x+3}{2} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+16x+15=4\times \frac{\left(2x+3\right)\left(2x+5\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}+16x+15=4\times \frac{\left(2x+3\right)\left(2x+5\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}+16x+15=\left(2x+3\right)\left(2x+5\right)$$

Evaluate

$$\left(2x+3\right)\left(2x+5\right)$$