Solve 7x^2-74x-120

$$7 { x }^{ 2 } -74x-120$$

Factor

$$\left(x-12\right)\left(7x+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 7x^{2}+ax+bx-120. To find a and b, set up a system to be solved.

$$a+b=-74 ab=7\left(-120\right)=-840$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -840.

$$1,-840 2,-420 3,-280 4,-210 5,-168 6,-140 7,-120 8,-105 10,-84 12,-70 14,-60 15,-56 20,-42 21,-40 24,-35 28,-30$$

Calculate the sum for each pair.

$$1-840=-839 2-420=-418 3-280=-277 4-210=-206 5-168=-163 6-140=-134 7-120=-113 8-105=-97 10-84=-74 12-70=-58 14-60=-46 15-56=-41 20-42=-22 21-40=-19 24-35=-11 28-30=-2$$

The solution is the pair that gives sum -74.

$$a=-84 b=10$$

Rewrite 7x^{2}-74x-120 as \left(7x^{2}-84x\right)+\left(10x-120\right).

$$\left(7x^{2}-84x\right)+\left(10x-120\right)$$

Factor out 7x in the first and 10 in the second group.

$$7x\left(x-12\right)+10\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(7x+10\right)$$

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.

$$7x^{2}-74x-120=0$$

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.

$$x=\frac{-\left(-74\right)±\sqrt{\left(-74\right)^{2}-4\times 7\left(-120\right)}}{2\times 7}$$

Square -74.

$$x=\frac{-\left(-74\right)±\sqrt{5476-4\times 7\left(-120\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-74\right)±\sqrt{5476-28\left(-120\right)}}{2\times 7}$$

Multiply -28 times -120.

$$x=\frac{-\left(-74\right)±\sqrt{5476+3360}}{2\times 7}$$

Add 5476 to 3360.

$$x=\frac{-\left(-74\right)±\sqrt{8836}}{2\times 7}$$

Take the square root of 8836.

$$x=\frac{-\left(-74\right)±94}{2\times 7}$$

The opposite of -74 is 74.

$$x=\frac{74±94}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{74±94}{14}$$

Now solve the equation x=\frac{74±94}{14} when ± is plus. Add 74 to 94.

$$x=\frac{168}{14}$$

Divide 168 by 14.

$$x=12$$

Now solve the equation x=\frac{74±94}{14} when ± is minus. Subtract 94 from 74.

$$x=-\frac{20}{14}$$

Reduce the fraction \frac{-20}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and -\frac{10}{7} for x_{2}.

$$7x^{2}-74x-120=7\left(x-12\right)\left(x-\left(-\frac{10}{7}\right)\right)$$

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

$$7x^{2}-74x-120=7\left(x-12\right)\left(x+\frac{10}{7}\right)$$

Add \frac{10}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$7x^{2}-74x-120=7\left(x-12\right)\times \frac{7x+10}{7}$$

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

$$7x^{2}-74x-120=\left(x-12\right)\left(7x+10\right)$$

Evaluate

$$\left(x-12\right)\left(7x+10\right)$$