Solve 15x^2+28x+12

$$15 x ^ { 2 } + 28 x + 12$$

Factor

$$\left(3x+2\right)\left(5x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=28 ab=15\times 12=180$$

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 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 28.

$$a=10 b=18$$

Rewrite 15x^{2}+28x+12 as \left(15x^{2}+10x\right)+\left(18x+12\right).

$$\left(15x^{2}+10x\right)+\left(18x+12\right)$$

Factor out 5x in the first and 6 in the second group.

$$5x\left(3x+2\right)+6\left(3x+2\right)$$

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

$$\left(3x+2\right)\left(5x+6\right)$$

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.

$$15x^{2}+28x+12=0$$

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.

$$x=\frac{-28±\sqrt{28^{2}-4\times 15\times 12}}{2\times 15}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 15\times 12}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-28±\sqrt{784-60\times 12}}{2\times 15}$$

Multiply -60 times 12.

$$x=\frac{-28±\sqrt{784-720}}{2\times 15}$$

Add 784 to -720.

$$x=\frac{-28±\sqrt{64}}{2\times 15}$$

Take the square root of 64.

$$x=\frac{-28±8}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-28±8}{30}$$

Now solve the equation x=\frac{-28±8}{30} when ± is plus. Add -28 to 8.

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

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

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

Now solve the equation x=\frac{-28±8}{30} when ± is minus. Subtract 8 from -28.

$$x=-\frac{36}{30}$$

Reduce the fraction \frac{-36}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{6}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3} for x_{1} and -\frac{6}{5} for x_{2}.

$$15x^{2}+28x+12=15\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-\frac{6}{5}\right)\right)$$

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

$$15x^{2}+28x+12=15\left(x+\frac{2}{3}\right)\left(x+\frac{6}{5}\right)$$

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

$$15x^{2}+28x+12=15\times \frac{3x+2}{3}\left(x+\frac{6}{5}\right)$$

Add \frac{6}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+28x+12=15\times \frac{3x+2}{3}\times \frac{5x+6}{5}$$

Multiply \frac{3x+2}{3} times \frac{5x+6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+28x+12=15\times \frac{\left(3x+2\right)\left(5x+6\right)}{3\times 5}$$

Multiply 3 times 5.

$$15x^{2}+28x+12=15\times \frac{\left(3x+2\right)\left(5x+6\right)}{15}$$

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

$$15x^{2}+28x+12=\left(3x+2\right)\left(5x+6\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 15

$$x ^ 2 +\frac{28}{15}x +\frac{4}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{28}{15} rs = \frac{4}{5}$$

Two numbers r and s sum up to -\frac{28}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{28}{15} = -\frac{14}{15}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{14}{15} - u s = -\frac{14}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{5}

$$(-\frac{14}{15} - u) (-\frac{14}{15} + u) = \frac{4}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{196}{225} - u^2 = \frac{4}{5}$$

Simplify the expression by subtracting \frac{196}{225} on both sides

$$-u^2 = \frac{4}{5}-\frac{196}{225} = -\frac{16}{225}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{16}{225} u = \pm\sqrt{\frac{16}{225}} = \pm \frac{4}{15}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{14}{15} - \frac{4}{15} = -1.200 s = -\frac{14}{15} + \frac{4}{15} = -0.667$$

Evaluate

$$\left(3x+2\right)\left(5x+6\right)$$