Solve 6x^2+7x-24

$$6 x ^ { 2 } + 7 x - 24$$

Factor

$$\left(2x-3\right)\left(3x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=7 ab=6\left(-24\right)=-144$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 7.

$$a=-9 b=16$$

Rewrite 6x^{2}+7x-24 as \left(6x^{2}-9x\right)+\left(16x-24\right).

$$\left(6x^{2}-9x\right)+\left(16x-24\right)$$

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

$$3x\left(2x-3\right)+8\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(3x+8\right)$$

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.

$$6x^{2}+7x-24=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 6\left(-24\right)}}{2\times 6}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 6\left(-24\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-7±\sqrt{49-24\left(-24\right)}}{2\times 6}$$

Multiply -24 times -24.

$$x=\frac{-7±\sqrt{49+576}}{2\times 6}$$

Add 49 to 576.

$$x=\frac{-7±\sqrt{625}}{2\times 6}$$

Take the square root of 625.

$$x=\frac{-7±25}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-7±25}{12}$$

Now solve the equation x=\frac{-7±25}{12} when ± is plus. Add -7 to 25.

$$x=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-7±25}{12} when ± is minus. Subtract 25 from -7.

$$x=-\frac{32}{12}$$

Reduce the fraction \frac{-32}{12} to lowest terms by extracting and canceling out 4.

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

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

$$6x^{2}+7x-24=6\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{8}{3}\right)\right)$$

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

$$6x^{2}+7x-24=6\left(x-\frac{3}{2}\right)\left(x+\frac{8}{3}\right)$$

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

$$6x^{2}+7x-24=6\times \frac{2x-3}{2}\left(x+\frac{8}{3}\right)$$

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

$$6x^{2}+7x-24=6\times \frac{2x-3}{2}\times \frac{3x+8}{3}$$

Multiply \frac{2x-3}{2} times \frac{3x+8}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+7x-24=6\times \frac{\left(2x-3\right)\left(3x+8\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}+7x-24=6\times \frac{\left(2x-3\right)\left(3x+8\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}+7x-24=\left(2x-3\right)\left(3x+8\right)$$

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 6

$$x ^ 2 +\frac{7}{6}x -4 = 0$$

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

$$r + s = -\frac{7}{6} rs = -4$$

Two numbers r and s sum up to -\frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{6} = -\frac{7}{12}. 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>

$$r = -\frac{7}{12} - u s = -\frac{7}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-\frac{7}{12} - u) (-\frac{7}{12} + u) = -4$$

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

$$\frac{49}{144} - u^2 = -4$$

Simplify the expression by subtracting \frac{49}{144} on both sides

$$-u^2 = -4-\frac{49}{144} = -\frac{625}{144}$$

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

$$u^2 = \frac{625}{144} u = \pm\sqrt{\frac{625}{144}} = \pm \frac{25}{12}$$

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

$$r =-\frac{7}{12} - \frac{25}{12} = -2.667 s = -\frac{7}{12} + \frac{25}{12} = 1.500$$

Evaluate

$$\left(2x-3\right)\left(3x+8\right)$$