Solve 7x^2+17x-12

$$7 { x }^{ 2 } +17x-12$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 7x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=17 ab=7\left(-12\right)=-84$$

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 -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 17.

$$a=-4 b=21$$

Rewrite 7x^{2}+17x-12 as \left(7x^{2}-4x\right)+\left(21x-12\right).

$$\left(7x^{2}-4x\right)+\left(21x-12\right)$$

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

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

Factor out common term 7x-4 by using distributive property.

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

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.

$$7x^{2}+17x-12=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 7\left(-12\right)}}{2\times 7}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 7\left(-12\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-17±\sqrt{289-28\left(-12\right)}}{2\times 7}$$

Multiply -28 times -12.

$$x=\frac{-17±\sqrt{289+336}}{2\times 7}$$

Add 289 to 336.

$$x=\frac{-17±\sqrt{625}}{2\times 7}$$

Take the square root of 625.

$$x=\frac{-17±25}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-17±25}{14}$$

Now solve the equation x=\frac{-17±25}{14} when ± is plus. Add -17 to 25.

$$x=\frac{8}{14}$$

Reduce the fraction \frac{8}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{7}$$

Now solve the equation x=\frac{-17±25}{14} when ± is minus. Subtract 25 from -17.

$$x=-\frac{42}{14}$$

Divide -42 by 14.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{7} for x_{1} and -3 for x_{2}.

$$7x^{2}+17x-12=7\left(x-\frac{4}{7}\right)\left(x-\left(-3\right)\right)$$

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

$$7x^{2}+17x-12=7\left(x-\frac{4}{7}\right)\left(x+3\right)$$

Subtract \frac{4}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$7x^{2}+17x-12=7\times \frac{7x-4}{7}\left(x+3\right)$$

Cancel out 7, the greatest common factor in 7 and 7.

$$7x^{2}+17x-12=\left(7x-4\right)\left(x+3\right)$$

Evaluate

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