Solve 16x^2+34x-15=

$$16 x ^ { 2 } + 34 x - 15 =$$

Factor

$$\left(8x-3\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=34 ab=16\left(-15\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 34.

$$a=-6 b=40$$

Rewrite 16x^{2}+34x-15 as \left(16x^{2}-6x\right)+\left(40x-15\right).

$$\left(16x^{2}-6x\right)+\left(40x-15\right)$$

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

$$2x\left(8x-3\right)+5\left(8x-3\right)$$

Factor out common term 8x-3 by using distributive property.

$$\left(8x-3\right)\left(2x+5\right)$$

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.

$$16x^{2}+34x-15=0$$

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.

$$x=\frac{-34±\sqrt{34^{2}-4\times 16\left(-15\right)}}{2\times 16}$$

Square 34.

$$x=\frac{-34±\sqrt{1156-4\times 16\left(-15\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-34±\sqrt{1156-64\left(-15\right)}}{2\times 16}$$

Multiply -64 times -15.

$$x=\frac{-34±\sqrt{1156+960}}{2\times 16}$$

Add 1156 to 960.

$$x=\frac{-34±\sqrt{2116}}{2\times 16}$$

Take the square root of 2116.

$$x=\frac{-34±46}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-34±46}{32}$$

Now solve the equation x=\frac{-34±46}{32} when ± is plus. Add -34 to 46.

$$x=\frac{12}{32}$$

Reduce the fraction \frac{12}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{8}$$

Now solve the equation x=\frac{-34±46}{32} when ± is minus. Subtract 46 from -34.

$$x=-\frac{80}{32}$$

Reduce the fraction \frac{-80}{32} to lowest terms by extracting and canceling out 16.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{8} for x_{1} and -\frac{5}{2} for x_{2}.

$$16x^{2}+34x-15=16\left(x-\frac{3}{8}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$16x^{2}+34x-15=16\left(x-\frac{3}{8}\right)\left(x+\frac{5}{2}\right)$$

Subtract \frac{3}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+34x-15=16\times \frac{8x-3}{8}\left(x+\frac{5}{2}\right)$$

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

$$16x^{2}+34x-15=16\times \frac{8x-3}{8}\times \frac{2x+5}{2}$$

Multiply \frac{8x-3}{8} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+34x-15=16\times \frac{\left(8x-3\right)\left(2x+5\right)}{8\times 2}$$

Multiply 8 times 2.

$$16x^{2}+34x-15=16\times \frac{\left(8x-3\right)\left(2x+5\right)}{16}$$

Cancel out 16, the greatest common factor in 16 and 16.

$$16x^{2}+34x-15=\left(8x-3\right)\left(2x+5\right)$$

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 16

$$x ^ 2 +\frac{17}{8}x -\frac{15}{16} = 0$$

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

$$r + s = -\frac{17}{8} rs = -\frac{15}{16}$$

Two numbers r and s sum up to -\frac{17}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{8} = -\frac{17}{16}. 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>

$$r = -\frac{17}{16} - u s = -\frac{17}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{15}{16}

$$(-\frac{17}{16} - u) (-\frac{17}{16} + u) = -\frac{15}{16}$$

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

$$\frac{289}{256} - u^2 = -\frac{15}{16}$$

Simplify the expression by subtracting \frac{289}{256} on both sides

$$-u^2 = -\frac{15}{16}-\frac{289}{256} = -\frac{529}{256}$$

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

$$u^2 = \frac{529}{256} u = \pm\sqrt{\frac{529}{256}} = \pm \frac{23}{16}$$

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

$$r =-\frac{17}{16} - \frac{23}{16} = -2.500 s = -\frac{17}{16} + \frac{23}{16} = 0.375$$

Evaluate

$$\left(8x-3\right)\left(2x+5\right)$$