Solve 7x^2-26x-45

$$7 { x }^{ 2 } -26x-45$$

Factor

$$\left(x-5\right)\left(7x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 7x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=-26 ab=7\left(-45\right)=-315$$

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 -315.

$$1,-315 3,-105 5,-63 7,-45 9,-35 15,-21$$

Calculate the sum for each pair.

$$1-315=-314 3-105=-102 5-63=-58 7-45=-38 9-35=-26 15-21=-6$$

The solution is the pair that gives sum -26.

$$a=-35 b=9$$

Rewrite 7x^{2}-26x-45 as \left(7x^{2}-35x\right)+\left(9x-45\right).

$$\left(7x^{2}-35x\right)+\left(9x-45\right)$$

Factor out 7x in the first and 9 in the second group.

$$7x\left(x-5\right)+9\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(7x+9\right)$$

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.

$$7x^{2}-26x-45=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 7\left(-45\right)}}{2\times 7}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 7\left(-45\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-26\right)±\sqrt{676-28\left(-45\right)}}{2\times 7}$$

Multiply -28 times -45.

$$x=\frac{-\left(-26\right)±\sqrt{676+1260}}{2\times 7}$$

Add 676 to 1260.

$$x=\frac{-\left(-26\right)±\sqrt{1936}}{2\times 7}$$

Take the square root of 1936.

$$x=\frac{-\left(-26\right)±44}{2\times 7}$$

The opposite of -26 is 26.

$$x=\frac{26±44}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{26±44}{14}$$

Now solve the equation x=\frac{26±44}{14} when ± is plus. Add 26 to 44.

$$x=\frac{70}{14}$$

Divide 70 by 14.

$$x=5$$

Now solve the equation x=\frac{26±44}{14} when ± is minus. Subtract 44 from 26.

$$x=-\frac{18}{14}$$

Reduce the fraction \frac{-18}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -\frac{9}{7} for x_{2}.

$$7x^{2}-26x-45=7\left(x-5\right)\left(x-\left(-\frac{9}{7}\right)\right)$$

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

$$7x^{2}-26x-45=7\left(x-5\right)\left(x+\frac{9}{7}\right)$$

Add \frac{9}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$7x^{2}-26x-45=7\left(x-5\right)\times \frac{7x+9}{7}$$

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

$$7x^{2}-26x-45=\left(x-5\right)\left(7x+9\right)$$

Evaluate

$$\left(x-5\right)\left(7x+9\right)$$