Solve 3x^2+10x-48

$$3 { x }^{ 2 } +10x-48$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\left(-48\right)=-144$$

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 10.

$$a=-8 b=18$$

Rewrite 3x^{2}+10x-48 as \left(3x^{2}-8x\right)+\left(18x-48\right).

$$\left(3x^{2}-8x\right)+\left(18x-48\right)$$

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

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

Factor out common term 3x-8 by using distributive property.

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

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.

$$3x^{2}+10x-48=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\left(-48\right)}}{2\times 3}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 3\left(-48\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-10±\sqrt{100-12\left(-48\right)}}{2\times 3}$$

Multiply -12 times -48.

$$x=\frac{-10±\sqrt{100+576}}{2\times 3}$$

Add 100 to 576.

$$x=\frac{-10±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$x=\frac{-10±26}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-10±26}{6}$$

Now solve the equation x=\frac{-10±26}{6} when ± is plus. Add -10 to 26.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-10±26}{6} when ± is minus. Subtract 26 from -10.

$$x=-\frac{36}{6}$$

Divide -36 by 6.

$$x=-6$$

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

$$3x^{2}+10x-48=3\left(x-\frac{8}{3}\right)\left(x-\left(-6\right)\right)$$

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

$$3x^{2}+10x-48=3\left(x-\frac{8}{3}\right)\left(x+6\right)$$

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

$$3x^{2}+10x-48=3\times \frac{3x-8}{3}\left(x+6\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}+10x-48=\left(3x-8\right)\left(x+6\right)$$

Evaluate

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