Solve 16x^2+10x-9=

$$16 x ^ { 2 } + 10 x - 9 =$$

Factor

$$\left(2x-1\right)\left(8x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=10 ab=16\left(-9\right)=-144$$

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 -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 10.

$$a=-8 b=18$$

Rewrite 16x^{2}+10x-9 as \left(16x^{2}-8x\right)+\left(18x-9\right).

$$\left(16x^{2}-8x\right)+\left(18x-9\right)$$

Factor out 8x in the first and 9 in the second group.

$$8x\left(2x-1\right)+9\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(8x+9\right)$$

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.

$$16x^{2}+10x-9=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 16\left(-9\right)}}{2\times 16}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 16\left(-9\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-10±\sqrt{100-64\left(-9\right)}}{2\times 16}$$

Multiply -64 times -9.

$$x=\frac{-10±\sqrt{100+576}}{2\times 16}$$

Add 100 to 576.

$$x=\frac{-10±\sqrt{676}}{2\times 16}$$

Take the square root of 676.

$$x=\frac{-10±26}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-10±26}{32}$$

Now solve the equation x=\frac{-10±26}{32} when ± is plus. Add -10 to 26.

$$x=\frac{16}{32}$$

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

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-10±26}{32} when ± is minus. Subtract 26 from -10.

$$x=-\frac{36}{32}$$

Reduce the fraction \frac{-36}{32} to lowest terms by extracting and canceling out 4.

$$x=-\frac{9}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and -\frac{9}{8} for x_{2}.

$$16x^{2}+10x-9=16\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{9}{8}\right)\right)$$

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

$$16x^{2}+10x-9=16\left(x-\frac{1}{2}\right)\left(x+\frac{9}{8}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+10x-9=16\times \frac{2x-1}{2}\left(x+\frac{9}{8}\right)$$

Add \frac{9}{8} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+10x-9=16\times \frac{2x-1}{2}\times \frac{8x+9}{8}$$

Multiply \frac{2x-1}{2} times \frac{8x+9}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+10x-9=16\times \frac{\left(2x-1\right)\left(8x+9\right)}{2\times 8}$$

Multiply 2 times 8.

$$16x^{2}+10x-9=16\times \frac{\left(2x-1\right)\left(8x+9\right)}{16}$$

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

$$16x^{2}+10x-9=\left(2x-1\right)\left(8x+9\right)$$

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

$$x ^ 2 +\frac{5}{8}x -\frac{9}{16} = 0$$

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

$$r + s = -\frac{5}{8} rs = -\frac{9}{16}$$

Two numbers r and s sum up to -\frac{5}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{8} = -\frac{5}{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>

$$r = -\frac{5}{16} - u s = -\frac{5}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{16}

$$(-\frac{5}{16} - u) (-\frac{5}{16} + u) = -\frac{9}{16}$$

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

$$\frac{25}{256} - u^2 = -\frac{9}{16}$$

Simplify the expression by subtracting \frac{25}{256} on both sides

$$-u^2 = -\frac{9}{16}-\frac{25}{256} = -\frac{169}{256}$$

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

$$u^2 = \frac{169}{256} u = \pm\sqrt{\frac{169}{256}} = \pm \frac{13}{16}$$

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

$$r =-\frac{5}{16} - \frac{13}{16} = -1.125 s = -\frac{5}{16} + \frac{13}{16} = 0.500$$

Evaluate

$$\left(2x-1\right)\left(8x+9\right)$$