Solve 14x^2+9x-65=

$$14 { x }^{ 2 } +9x-65=$$

Factor

$$\left(7x-13\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 14x^{2}+ax+bx-65. To find a and b, set up a system to be solved.

$$a+b=9 ab=14\left(-65\right)=-910$$

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 -910.

$$-1,910 -2,455 -5,182 -7,130 -10,91 -13,70 -14,65 -26,35$$

Calculate the sum for each pair.

$$-1+910=909 -2+455=453 -5+182=177 -7+130=123 -10+91=81 -13+70=57 -14+65=51 -26+35=9$$

The solution is the pair that gives sum 9.

$$a=-26 b=35$$

Rewrite 14x^{2}+9x-65 as \left(14x^{2}-26x\right)+\left(35x-65\right).

$$\left(14x^{2}-26x\right)+\left(35x-65\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(7x-13\right)+5\left(7x-13\right)$$

Factor out common term 7x-13 by using distributive property.

$$\left(7x-13\right)\left(2x+5\right)$$

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.

$$14x^{2}+9x-65=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 14\left(-65\right)}}{2\times 14}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 14\left(-65\right)}}{2\times 14}$$

Multiply -4 times 14.

$$x=\frac{-9±\sqrt{81-56\left(-65\right)}}{2\times 14}$$

Multiply -56 times -65.

$$x=\frac{-9±\sqrt{81+3640}}{2\times 14}$$

Add 81 to 3640.

$$x=\frac{-9±\sqrt{3721}}{2\times 14}$$

Take the square root of 3721.

$$x=\frac{-9±61}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{-9±61}{28}$$

Now solve the equation x=\frac{-9±61}{28} when ± is plus. Add -9 to 61.

$$x=\frac{52}{28}$$

Reduce the fraction \frac{52}{28} to lowest terms by extracting and canceling out 4.

$$x=\frac{13}{7}$$

Now solve the equation x=\frac{-9±61}{28} when ± is minus. Subtract 61 from -9.

$$x=-\frac{70}{28}$$

Reduce the fraction \frac{-70}{28} to lowest terms by extracting and canceling out 14.

$$x=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13}{7} for x_{1} and -\frac{5}{2} for x_{2}.

$$14x^{2}+9x-65=14\left(x-\frac{13}{7}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$14x^{2}+9x-65=14\left(x-\frac{13}{7}\right)\left(x+\frac{5}{2}\right)$$

Subtract \frac{13}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14x^{2}+9x-65=14\times \frac{7x-13}{7}\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$14x^{2}+9x-65=14\times \frac{7x-13}{7}\times \frac{2x+5}{2}$$

Multiply \frac{7x-13}{7} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$14x^{2}+9x-65=14\times \frac{\left(7x-13\right)\left(2x+5\right)}{7\times 2}$$

Multiply 7 times 2.

$$14x^{2}+9x-65=14\times \frac{\left(7x-13\right)\left(2x+5\right)}{14}$$

Cancel out 14, the greatest common factor in 14 and 14.

$$14x^{2}+9x-65=\left(7x-13\right)\left(2x+5\right)$$

Evaluate

$$\left(7x-13\right)\left(2x+5\right)$$