Solve 15x^2+31x+10

$$15 { x }^{ 2 } +31x+10$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=31 ab=15\times 10=150$$

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 150.

$$1,150 2,75 3,50 5,30 6,25 10,15$$

Calculate the sum for each pair.

$$1+150=151 2+75=77 3+50=53 5+30=35 6+25=31 10+15=25$$

The solution is the pair that gives sum 31.

$$a=6 b=25$$

Rewrite 15x^{2}+31x+10 as \left(15x^{2}+6x\right)+\left(25x+10\right).

$$\left(15x^{2}+6x\right)+\left(25x+10\right)$$

Factor out 3x in the first and 5 in the second group.

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

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

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

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.

$$15x^{2}+31x+10=0$$

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.

$$x=\frac{-31±\sqrt{31^{2}-4\times 15\times 10}}{2\times 15}$$

Square 31.

$$x=\frac{-31±\sqrt{961-4\times 15\times 10}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-31±\sqrt{961-60\times 10}}{2\times 15}$$

Multiply -60 times 10.

$$x=\frac{-31±\sqrt{961-600}}{2\times 15}$$

Add 961 to -600.

$$x=\frac{-31±\sqrt{361}}{2\times 15}$$

Take the square root of 361.

$$x=\frac{-31±19}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-31±19}{30}$$

Now solve the equation x=\frac{-31±19}{30} when ± is plus. Add -31 to 19.

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

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{-31±19}{30} when ± is minus. Subtract 19 from -31.

$$x=-\frac{50}{30}$$

Reduce the fraction \frac{-50}{30} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{3}$$

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{5}{3} for x_{2}.

$$15x^{2}+31x+10=15\left(x-\left(-\frac{2}{5}\right)\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$15x^{2}+31x+10=15\left(x+\frac{2}{5}\right)\left(x+\frac{5}{3}\right)$$

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

$$15x^{2}+31x+10=15\times \frac{5x+2}{5}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+31x+10=15\times \frac{5x+2}{5}\times \frac{3x+5}{3}$$

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

$$15x^{2}+31x+10=15\times \frac{\left(5x+2\right)\left(3x+5\right)}{5\times 3}$$

Multiply 5 times 3.

$$15x^{2}+31x+10=15\times \frac{\left(5x+2\right)\left(3x+5\right)}{15}$$

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

$$15x^{2}+31x+10=\left(5x+2\right)\left(3x+5\right)$$

Evaluate

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