Solve 24x^2+10x-21

$$24 { x }^{ 2 } +10x-21$$

Factor

$$\left(4x-3\right)\left(6x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=10 ab=24\left(-21\right)=-504$$

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 -504.

$$-1,504 -2,252 -3,168 -4,126 -6,84 -7,72 -8,63 -9,56 -12,42 -14,36 -18,28 -21,24$$

Calculate the sum for each pair.

$$-1+504=503 -2+252=250 -3+168=165 -4+126=122 -6+84=78 -7+72=65 -8+63=55 -9+56=47 -12+42=30 -14+36=22 -18+28=10 -21+24=3$$

The solution is the pair that gives sum 10.

$$a=-18 b=28$$

Rewrite 24x^{2}+10x-21 as \left(24x^{2}-18x\right)+\left(28x-21\right).

$$\left(24x^{2}-18x\right)+\left(28x-21\right)$$

Factor out 6x in the first and 7 in the second group.

$$6x\left(4x-3\right)+7\left(4x-3\right)$$

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

$$\left(4x-3\right)\left(6x+7\right)$$

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.

$$24x^{2}+10x-21=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 24\left(-21\right)}}{2\times 24}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 24\left(-21\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-10±\sqrt{100-96\left(-21\right)}}{2\times 24}$$

Multiply -96 times -21.

$$x=\frac{-10±\sqrt{100+2016}}{2\times 24}$$

Add 100 to 2016.

$$x=\frac{-10±\sqrt{2116}}{2\times 24}$$

Take the square root of 2116.

$$x=\frac{-10±46}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-10±46}{48}$$

Now solve the equation x=\frac{-10±46}{48} when ± is plus. Add -10 to 46.

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

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{-10±46}{48} when ± is minus. Subtract 46 from -10.

$$x=-\frac{56}{48}$$

Reduce the fraction \frac{-56}{48} to lowest terms by extracting and canceling out 8.

$$x=-\frac{7}{6}$$

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{7}{6} for x_{2}.

$$24x^{2}+10x-21=24\left(x-\frac{3}{4}\right)\left(x-\left(-\frac{7}{6}\right)\right)$$

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

$$24x^{2}+10x-21=24\left(x-\frac{3}{4}\right)\left(x+\frac{7}{6}\right)$$

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

$$24x^{2}+10x-21=24\times \frac{4x-3}{4}\left(x+\frac{7}{6}\right)$$

Add \frac{7}{6} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+10x-21=24\times \frac{4x-3}{4}\times \frac{6x+7}{6}$$

Multiply \frac{4x-3}{4} times \frac{6x+7}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+10x-21=24\times \frac{\left(4x-3\right)\left(6x+7\right)}{4\times 6}$$

Multiply 4 times 6.

$$24x^{2}+10x-21=24\times \frac{\left(4x-3\right)\left(6x+7\right)}{24}$$

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

$$24x^{2}+10x-21=\left(4x-3\right)\left(6x+7\right)$$

Evaluate

$$\left(4x-3\right)\left(6x+7\right)$$