Solve 15x^2+58x+48

$$15 x ^ { 2 } + 58 x + 48$$

Factor

$$\left(3x+8\right)\left(5x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=58 ab=15\times 48=720$$

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 720.

$$1,720 2,360 3,240 4,180 5,144 6,120 8,90 9,80 10,72 12,60 15,48 16,45 18,40 20,36 24,30$$

Calculate the sum for each pair.

$$1+720=721 2+360=362 3+240=243 4+180=184 5+144=149 6+120=126 8+90=98 9+80=89 10+72=82 12+60=72 15+48=63 16+45=61 18+40=58 20+36=56 24+30=54$$

The solution is the pair that gives sum 58.

$$a=18 b=40$$

Rewrite 15x^{2}+58x+48 as \left(15x^{2}+18x\right)+\left(40x+48\right).

$$\left(15x^{2}+18x\right)+\left(40x+48\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(5x+6\right)+8\left(5x+6\right)$$

Factor out common term 5x+6 by using distributive property.

$$\left(5x+6\right)\left(3x+8\right)$$

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.

$$15x^{2}+58x+48=0$$

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.

$$x=\frac{-58±\sqrt{58^{2}-4\times 15\times 48}}{2\times 15}$$

Square 58.

$$x=\frac{-58±\sqrt{3364-4\times 15\times 48}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-58±\sqrt{3364-60\times 48}}{2\times 15}$$

Multiply -60 times 48.

$$x=\frac{-58±\sqrt{3364-2880}}{2\times 15}$$

Add 3364 to -2880.

$$x=\frac{-58±\sqrt{484}}{2\times 15}$$

Take the square root of 484.

$$x=\frac{-58±22}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-58±22}{30}$$

Now solve the equation x=\frac{-58±22}{30} when ± is plus. Add -58 to 22.

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

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

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

Now solve the equation x=\frac{-58±22}{30} when ± is minus. Subtract 22 from -58.

$$x=-\frac{80}{30}$$

Reduce the fraction \frac{-80}{30} to lowest terms by extracting and canceling out 10.

$$x=-\frac{8}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{6}{5} for x_{1} and -\frac{8}{3} for x_{2}.

$$15x^{2}+58x+48=15\left(x-\left(-\frac{6}{5}\right)\right)\left(x-\left(-\frac{8}{3}\right)\right)$$

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

$$15x^{2}+58x+48=15\left(x+\frac{6}{5}\right)\left(x+\frac{8}{3}\right)$$

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

$$15x^{2}+58x+48=15\times \frac{5x+6}{5}\left(x+\frac{8}{3}\right)$$

Add \frac{8}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+58x+48=15\times \frac{5x+6}{5}\times \frac{3x+8}{3}$$

Multiply \frac{5x+6}{5} times \frac{3x+8}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+58x+48=15\times \frac{\left(5x+6\right)\left(3x+8\right)}{5\times 3}$$

Multiply 5 times 3.

$$15x^{2}+58x+48=15\times \frac{\left(5x+6\right)\left(3x+8\right)}{15}$$

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

$$15x^{2}+58x+48=\left(5x+6\right)\left(3x+8\right)$$

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

$$x ^ 2 +\frac{58}{15}x +\frac{16}{5} = 0$$

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

$$r + s = -\frac{58}{15} rs = \frac{16}{5}$$

Two numbers r and s sum up to -\frac{58}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{58}{15} = -\frac{29}{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>

$$r = -\frac{29}{15} - u s = -\frac{29}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{16}{5}

$$(-\frac{29}{15} - u) (-\frac{29}{15} + u) = \frac{16}{5}$$

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

$$\frac{841}{225} - u^2 = \frac{16}{5}$$

Simplify the expression by subtracting \frac{841}{225} on both sides

$$-u^2 = \frac{16}{5}-\frac{841}{225} = -\frac{121}{225}$$

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

$$u^2 = \frac{121}{225} u = \pm\sqrt{\frac{121}{225}} = \pm \frac{11}{15}$$

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

$$r =-\frac{29}{15} - \frac{11}{15} = -2.667 s = -\frac{29}{15} + \frac{11}{15} = -1.200$$

Evaluate

$$\left(3x+8\right)\left(5x+6\right)$$