Solve 15x^2-11x-14

$$15 x ^ { 2 } - 11 x - 14$$

Factor

$$\left(5x-7\right)\left(3x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-11 ab=15\left(-14\right)=-210$$

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 -210.

$$1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15$$

Calculate the sum for each pair.

$$1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1$$

The solution is the pair that gives sum -11.

$$a=-21 b=10$$

Rewrite 15x^{2}-11x-14 as \left(15x^{2}-21x\right)+\left(10x-14\right).

$$\left(15x^{2}-21x\right)+\left(10x-14\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(5x-7\right)+2\left(5x-7\right)$$

Factor out common term 5x-7 by using distributive property.

$$\left(5x-7\right)\left(3x+2\right)$$

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.

$$15x^{2}-11x-14=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 15\left(-14\right)}}{2\times 15}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 15\left(-14\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-11\right)±\sqrt{121-60\left(-14\right)}}{2\times 15}$$

Multiply -60 times -14.

$$x=\frac{-\left(-11\right)±\sqrt{121+840}}{2\times 15}$$

Add 121 to 840.

$$x=\frac{-\left(-11\right)±\sqrt{961}}{2\times 15}$$

Take the square root of 961.

$$x=\frac{-\left(-11\right)±31}{2\times 15}$$

The opposite of -11 is 11.

$$x=\frac{11±31}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{11±31}{30}$$

Now solve the equation x=\frac{11±31}{30} when ± is plus. Add 11 to 31.

$$x=\frac{42}{30}$$

Reduce the fraction \frac{42}{30} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{5}$$

Now solve the equation x=\frac{11±31}{30} when ± is minus. Subtract 31 from 11.

$$x=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

$$x=-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{5} for x_{1} and -\frac{2}{3} for x_{2}.

$$15x^{2}-11x-14=15\left(x-\frac{7}{5}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$15x^{2}-11x-14=15\left(x-\frac{7}{5}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{7}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-11x-14=15\times \frac{5x-7}{5}\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-11x-14=15\times \frac{5x-7}{5}\times \frac{3x+2}{3}$$

Multiply \frac{5x-7}{5} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-11x-14=15\times \frac{\left(5x-7\right)\left(3x+2\right)}{5\times 3}$$

Multiply 5 times 3.

$$15x^{2}-11x-14=15\times \frac{\left(5x-7\right)\left(3x+2\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15x^{2}-11x-14=\left(5x-7\right)\left(3x+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 15

$$x ^ 2 -\frac{11}{15}x -\frac{14}{15} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{11}{15} rs = -\frac{14}{15}$$

Two numbers r and s sum up to \frac{11}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{15} = \frac{11}{30}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{11}{30} - u s = \frac{11}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{14}{15}

$$(\frac{11}{30} - u) (\frac{11}{30} + u) = -\frac{14}{15}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{121}{900} - u^2 = -\frac{14}{15}$$

Simplify the expression by subtracting \frac{121}{900} on both sides

$$-u^2 = -\frac{14}{15}-\frac{121}{900} = -\frac{961}{900}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{961}{900} u = \pm\sqrt{\frac{961}{900}} = \pm \frac{31}{30}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{11}{30} - \frac{31}{30} = -0.667 s = \frac{11}{30} + \frac{31}{30} = 1.400$$

Evaluate

$$\left(5x-7\right)\left(3x+2\right)$$