Solve 40x^2-17x-12

$$40 x ^ { 2 } - 17 x - 12$$

Factor

$$\left(5x-4\right)\left(8x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 40x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-17 ab=40\left(-12\right)=-480$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -480.

$$1,-480 2,-240 3,-160 4,-120 5,-96 6,-80 8,-60 10,-48 12,-40 15,-32 16,-30 20,-24$$

Calculate the sum for each pair.

$$1-480=-479 2-240=-238 3-160=-157 4-120=-116 5-96=-91 6-80=-74 8-60=-52 10-48=-38 12-40=-28 15-32=-17 16-30=-14 20-24=-4$$

The solution is the pair that gives sum -17.

$$a=-32 b=15$$

Rewrite 40x^{2}-17x-12 as \left(40x^{2}-32x\right)+\left(15x-12\right).

$$\left(40x^{2}-32x\right)+\left(15x-12\right)$$

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

$$8x\left(5x-4\right)+3\left(5x-4\right)$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(8x+3\right)$$

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.

$$40x^{2}-17x-12=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 40\left(-12\right)}}{2\times 40}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 40\left(-12\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-17\right)±\sqrt{289-160\left(-12\right)}}{2\times 40}$$

Multiply -160 times -12.

$$x=\frac{-\left(-17\right)±\sqrt{289+1920}}{2\times 40}$$

Add 289 to 1920.

$$x=\frac{-\left(-17\right)±\sqrt{2209}}{2\times 40}$$

Take the square root of 2209.

$$x=\frac{-\left(-17\right)±47}{2\times 40}$$

The opposite of -17 is 17.

$$x=\frac{17±47}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{17±47}{80}$$

Now solve the equation x=\frac{17±47}{80} when ± is plus. Add 17 to 47.

$$x=\frac{64}{80}$$

Reduce the fraction \frac{64}{80} to lowest terms by extracting and canceling out 16.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{17±47}{80} when ± is minus. Subtract 47 from 17.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{5} for x_{1} and -\frac{3}{8} for x_{2}.

$$40x^{2}-17x-12=40\left(x-\frac{4}{5}\right)\left(x-\left(-\frac{3}{8}\right)\right)$$

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

$$40x^{2}-17x-12=40\left(x-\frac{4}{5}\right)\left(x+\frac{3}{8}\right)$$

Subtract \frac{4}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$40x^{2}-17x-12=40\times \frac{5x-4}{5}\left(x+\frac{3}{8}\right)$$

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

$$40x^{2}-17x-12=40\times \frac{5x-4}{5}\times \frac{8x+3}{8}$$

Multiply \frac{5x-4}{5} times \frac{8x+3}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$40x^{2}-17x-12=40\times \frac{\left(5x-4\right)\left(8x+3\right)}{5\times 8}$$

Multiply 5 times 8.

$$40x^{2}-17x-12=40\times \frac{\left(5x-4\right)\left(8x+3\right)}{40}$$

Cancel out 40, the greatest common factor in 40 and 40.

$$40x^{2}-17x-12=\left(5x-4\right)\left(8x+3\right)$$

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 40

$$x ^ 2 -\frac{17}{40}x -\frac{3}{10} = 0$$

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

$$r + s = \frac{17}{40} rs = -\frac{3}{10}$$

Two numbers r and s sum up to \frac{17}{40} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{40} = \frac{17}{80}. 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>

$$r = \frac{17}{80} - u s = \frac{17}{80} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}

$$(\frac{17}{80} - u) (\frac{17}{80} + u) = -\frac{3}{10}$$

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

$$\frac{289}{6400} - u^2 = -\frac{3}{10}$$

Simplify the expression by subtracting \frac{289}{6400} on both sides

$$-u^2 = -\frac{3}{10}-\frac{289}{6400} = -\frac{2209}{6400}$$

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

$$u^2 = \frac{2209}{6400} u = \pm\sqrt{\frac{2209}{6400}} = \pm \frac{47}{80}$$

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

$$r =\frac{17}{80} - \frac{47}{80} = -0.375 s = \frac{17}{80} + \frac{47}{80} = 0.800$$

Evaluate

$$\left(5x-4\right)\left(8x+3\right)$$