Solve 24x^2+38x+15

$$24 x ^ { 2 } + 38 x + 15$$

Factor

$$\left(4x+3\right)\left(6x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=38 ab=24\times 15=360$$

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 360.

$$1,360 2,180 3,120 4,90 5,72 6,60 8,45 9,40 10,36 12,30 15,24 18,20$$

Calculate the sum for each pair.

$$1+360=361 2+180=182 3+120=123 4+90=94 5+72=77 6+60=66 8+45=53 9+40=49 10+36=46 12+30=42 15+24=39 18+20=38$$

The solution is the pair that gives sum 38.

$$a=18 b=20$$

Rewrite 24x^{2}+38x+15 as \left(24x^{2}+18x\right)+\left(20x+15\right).

$$\left(24x^{2}+18x\right)+\left(20x+15\right)$$

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

$$6x\left(4x+3\right)+5\left(4x+3\right)$$

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

$$\left(4x+3\right)\left(6x+5\right)$$

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.

$$24x^{2}+38x+15=0$$

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.

$$x=\frac{-38±\sqrt{38^{2}-4\times 24\times 15}}{2\times 24}$$

Square 38.

$$x=\frac{-38±\sqrt{1444-4\times 24\times 15}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-38±\sqrt{1444-96\times 15}}{2\times 24}$$

Multiply -96 times 15.

$$x=\frac{-38±\sqrt{1444-1440}}{2\times 24}$$

Add 1444 to -1440.

$$x=\frac{-38±\sqrt{4}}{2\times 24}$$

Take the square root of 4.

$$x=\frac{-38±2}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-38±2}{48}$$

Now solve the equation x=\frac{-38±2}{48} when ± is plus. Add -38 to 2.

$$x=-\frac{36}{48}$$

Reduce the fraction \frac{-36}{48} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{-38±2}{48} when ± is minus. Subtract 2 from -38.

$$x=-\frac{40}{48}$$

Reduce the fraction \frac{-40}{48} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{4} for x_{1} and -\frac{5}{6} for x_{2}.

$$24x^{2}+38x+15=24\left(x-\left(-\frac{3}{4}\right)\right)\left(x-\left(-\frac{5}{6}\right)\right)$$

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

$$24x^{2}+38x+15=24\left(x+\frac{3}{4}\right)\left(x+\frac{5}{6}\right)$$

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

$$24x^{2}+38x+15=24\times \frac{4x+3}{4}\left(x+\frac{5}{6}\right)$$

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

$$24x^{2}+38x+15=24\times \frac{4x+3}{4}\times \frac{6x+5}{6}$$

Multiply \frac{4x+3}{4} times \frac{6x+5}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+38x+15=24\times \frac{\left(4x+3\right)\left(6x+5\right)}{4\times 6}$$

Multiply 4 times 6.

$$24x^{2}+38x+15=24\times \frac{\left(4x+3\right)\left(6x+5\right)}{24}$$

Cancel out 24, the greatest common factor in 24 and 24.

$$24x^{2}+38x+15=\left(4x+3\right)\left(6x+5\right)$$

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 24

$$x ^ 2 +\frac{19}{12}x +\frac{5}{8} = 0$$

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

$$r + s = -\frac{19}{12} rs = \frac{5}{8}$$

Two numbers r and s sum up to -\frac{19}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{12} = -\frac{19}{24}. 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>

$$r = -\frac{19}{24} - u s = -\frac{19}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{8}

$$(-\frac{19}{24} - u) (-\frac{19}{24} + u) = \frac{5}{8}$$

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

$$\frac{361}{576} - u^2 = \frac{5}{8}$$

Simplify the expression by subtracting \frac{361}{576} on both sides

$$-u^2 = \frac{5}{8}-\frac{361}{576} = -\frac{1}{576}$$

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

$$u^2 = \frac{1}{576} u = \pm\sqrt{\frac{1}{576}} = \pm \frac{1}{24}$$

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

$$r =-\frac{19}{24} - \frac{1}{24} = -0.833 s = -\frac{19}{24} + \frac{1}{24} = -0.750$$

Evaluate

$$\left(4x+3\right)\left(6x+5\right)$$