Solve 14x^2+11x-15

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

Factor

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

Solution Steps

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

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

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 -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.

$$a=-10 b=21$$

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

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

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

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

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

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

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.

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

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.

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

Square 11.

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

Multiply -4 times 14.

$$x=\frac{-11±\sqrt{121-56\left(-15\right)}}{2\times 14}$$

Multiply -56 times -15.

$$x=\frac{-11±\sqrt{121+840}}{2\times 14}$$

Add 121 to 840.

$$x=\frac{-11±\sqrt{961}}{2\times 14}$$

Take the square root of 961.

$$x=\frac{-11±31}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{-11±31}{28}$$

Now solve the equation x=\frac{-11±31}{28} when ± is plus. Add -11 to 31.

$$x=\frac{20}{28}$$

Reduce the fraction \frac{20}{28} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-11±31}{28} when ± is minus. Subtract 31 from -11.

$$x=-\frac{42}{28}$$

Reduce the fraction \frac{-42}{28} to lowest terms by extracting and canceling out 14.

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

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

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

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

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

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

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

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

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

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

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

Multiply 7 times 2.

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

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

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

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 14

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

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

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

Two numbers r and s sum up to -\frac{11}{14} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{14} = -\frac{11}{28}. 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>

$$r = -\frac{11}{28} - u s = -\frac{11}{28} + u$$

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

$$(-\frac{11}{28} - u) (-\frac{11}{28} + u) = -\frac{15}{14}$$

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

$$\frac{121}{784} - u^2 = -\frac{15}{14}$$

Simplify the expression by subtracting \frac{121}{784} on both sides

$$-u^2 = -\frac{15}{14}-\frac{121}{784} = -\frac{961}{784}$$

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

$$u^2 = \frac{961}{784} u = \pm\sqrt{\frac{961}{784}} = \pm \frac{31}{28}$$

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

$$r =-\frac{11}{28} - \frac{31}{28} = -1.500 s = -\frac{11}{28} + \frac{31}{28} = 0.714$$

Evaluate

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