Solve 15x^2+16x+4

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

Factor

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

Solution Steps

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

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

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

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

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

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

Factor out common term 5x+2 by using distributive property.

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

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.

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

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.

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

Square 16.

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

Multiply -4 times 15.

$$x=\frac{-16±\sqrt{256-60\times 4}}{2\times 15}$$

Multiply -60 times 4.

$$x=\frac{-16±\sqrt{256-240}}{2\times 15}$$

Add 256 to -240.

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

Take the square root of 16.

$$x=\frac{-16±4}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-16±4}{30}$$

Now solve the equation x=\frac{-16±4}{30} when ± is plus. Add -16 to 4.

$$x=-\frac{12}{30}$$

Reduce the fraction \frac{-12}{30} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-16±4}{30} when ± is minus. Subtract 4 from -16.

$$x=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

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

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

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

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

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

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

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

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

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

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

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

Multiply 5 times 3.

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

Cancel out 15, the greatest common factor in 15 and 15.

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

Evaluate

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