Solve 2x^2-9x-56

$$2 x ^ { 2 } - 9 x - 56$$

Factor

$$\left(x-8\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\left(-56\right)=-112$$

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 -112.

$$1,-112 2,-56 4,-28 7,-16 8,-14$$

Calculate the sum for each pair.

$$1-112=-111 2-56=-54 4-28=-24 7-16=-9 8-14=-6$$

The solution is the pair that gives sum -9.

$$a=-16 b=7$$

Rewrite 2x^{2}-9x-56 as \left(2x^{2}-16x\right)+\left(7x-56\right).

$$\left(2x^{2}-16x\right)+\left(7x-56\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-8\right)+7\left(x-8\right)$$

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

$$\left(x-8\right)\left(2x+7\right)$$

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.

$$2x^{2}-9x-56=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-56\right)}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-56\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-56\right)}}{2\times 2}$$

Multiply -8 times -56.

$$x=\frac{-\left(-9\right)±\sqrt{81+448}}{2\times 2}$$

Add 81 to 448.

$$x=\frac{-\left(-9\right)±\sqrt{529}}{2\times 2}$$

Take the square root of 529.

$$x=\frac{-\left(-9\right)±23}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±23}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±23}{4}$$

Now solve the equation x=\frac{9±23}{4} when ± is plus. Add 9 to 23.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{9±23}{4} when ± is minus. Subtract 23 from 9.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -\frac{7}{2} for x_{2}.

$$2x^{2}-9x-56=2\left(x-8\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

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

$$2x^{2}-9x-56=2\left(x-8\right)\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-9x-56=2\left(x-8\right)\times \frac{2x+7}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2x^{2}-9x-56=\left(x-8\right)\left(2x+7\right)$$

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 2

$$x ^ 2 -\frac{9}{2}x -28 = 0$$

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

$$r + s = \frac{9}{2} rs = -28$$

Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{4}. 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>

$$r = \frac{9}{4} - u s = \frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -28

$$(\frac{9}{4} - u) (\frac{9}{4} + u) = -28$$

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

$$\frac{81}{16} - u^2 = -28$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = -28-\frac{81}{16} = -\frac{529}{16}$$

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

$$u^2 = \frac{529}{16} u = \pm\sqrt{\frac{529}{16}} = \pm \frac{23}{4}$$

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

$$r =\frac{9}{4} - \frac{23}{4} = -3.500 s = \frac{9}{4} + \frac{23}{4} = 8$$

Evaluate

$$\left(x-8\right)\left(2x+7\right)$$