Solve 184x^2+140x-116

$$184 { x }^{ 2 } +140x-116$$

Factor

$$4\left(2x-1\right)\left(23x+29\right)$$

Solution Steps

Factor out 4.

$$4\left(46x^{2}+35x-29\right)$$

Consider 46x^{2}+35x-29. Factor the expression by grouping. First, the expression needs to be rewritten as 46x^{2}+ax+bx-29. To find a and b, set up a system to be solved.

$$a+b=35 ab=46\left(-29\right)=-1334$$

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 -1334.

$$-1,1334 -2,667 -23,58 -29,46$$

Calculate the sum for each pair.

$$-1+1334=1333 -2+667=665 -23+58=35 -29+46=17$$

The solution is the pair that gives sum 35.

$$a=-23 b=58$$

Rewrite 46x^{2}+35x-29 as \left(46x^{2}-23x\right)+\left(58x-29\right).

$$\left(46x^{2}-23x\right)+\left(58x-29\right)$$

Factor out 23x in the first and 29 in the second group.

$$23x\left(2x-1\right)+29\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(23x+29\right)$$

Rewrite the complete factored expression.

$$4\left(2x-1\right)\left(23x+29\right)$$

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.

$$184x^{2}+140x-116=0$$

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.

$$x=\frac{-140±\sqrt{140^{2}-4\times 184\left(-116\right)}}{2\times 184}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\times 184\left(-116\right)}}{2\times 184}$$

Multiply -4 times 184.

$$x=\frac{-140±\sqrt{19600-736\left(-116\right)}}{2\times 184}$$

Multiply -736 times -116.

$$x=\frac{-140±\sqrt{19600+85376}}{2\times 184}$$

Add 19600 to 85376.

$$x=\frac{-140±\sqrt{104976}}{2\times 184}$$

Take the square root of 104976.

$$x=\frac{-140±324}{2\times 184}$$

Multiply 2 times 184.

$$x=\frac{-140±324}{368}$$

Now solve the equation x=\frac{-140±324}{368} when ± is plus. Add -140 to 324.

$$x=\frac{184}{368}$$

Reduce the fraction \frac{184}{368} to lowest terms by extracting and canceling out 184.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-140±324}{368} when ± is minus. Subtract 324 from -140.

$$x=-\frac{464}{368}$$

Reduce the fraction \frac{-464}{368} to lowest terms by extracting and canceling out 16.

$$x=-\frac{29}{23}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and -\frac{29}{23} for x_{2}.

$$184x^{2}+140x-116=184\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{29}{23}\right)\right)$$

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

$$184x^{2}+140x-116=184\left(x-\frac{1}{2}\right)\left(x+\frac{29}{23}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$184x^{2}+140x-116=184\times \frac{2x-1}{2}\left(x+\frac{29}{23}\right)$$

Add \frac{29}{23} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$184x^{2}+140x-116=184\times \frac{2x-1}{2}\times \frac{23x+29}{23}$$

Multiply \frac{2x-1}{2} times \frac{23x+29}{23} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$184x^{2}+140x-116=184\times \frac{\left(2x-1\right)\left(23x+29\right)}{2\times 23}$$

Multiply 2 times 23.

$$184x^{2}+140x-116=184\times \frac{\left(2x-1\right)\left(23x+29\right)}{46}$$

Cancel out 46, the greatest common factor in 184 and 46.

$$184x^{2}+140x-116=4\left(2x-1\right)\left(23x+29\right)$$

Evaluate

$$184x^{2}+140x-116$$