Solve 4x^2+6x-40

$$4 x ^ { 2 } + 6 x - 40$$

Factor

$$2\left(2x-5\right)\left(x+4\right)$$

Solution Steps

Factor out 2.

$$2\left(2x^{2}+3x-20\right)$$

Consider 2x^{2}+3x-20. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-20\right)=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite 2x^{2}+3x-20 as \left(2x^{2}-5x\right)+\left(8x-20\right).

$$\left(2x^{2}-5x\right)+\left(8x-20\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x-5\right)+4\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$2\left(2x-5\right)\left(x+4\right)$$

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.

$$4x^{2}+6x-40=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-40\right)}}{2\times 4}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-40\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\left(-40\right)}}{2\times 4}$$

Multiply -16 times -40.

$$x=\frac{-6±\sqrt{36+640}}{2\times 4}$$

Add 36 to 640.

$$x=\frac{-6±\sqrt{676}}{2\times 4}$$

Take the square root of 676.

$$x=\frac{-6±26}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±26}{8}$$

Now solve the equation x=\frac{-6±26}{8} when ± is plus. Add -6 to 26.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-6±26}{8} when ± is minus. Subtract 26 from -6.

$$x=-\frac{32}{8}$$

Divide -32 by 8.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -4 for x_{2}.

$$4x^{2}+6x-40=4\left(x-\frac{5}{2}\right)\left(x-\left(-4\right)\right)$$

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

$$4x^{2}+6x-40=4\left(x-\frac{5}{2}\right)\left(x+4\right)$$

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

$$4x^{2}+6x-40=4\times \frac{2x-5}{2}\left(x+4\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4x^{2}+6x-40=2\left(2x-5\right)\left(x+4\right)$$

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 4

$$x ^ 2 +\frac{3}{2}x -10 = 0$$

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

$$r + s = -\frac{3}{2} rs = -10$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{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>

$$r = -\frac{3}{4} - u s = -\frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -10$$

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

$$\frac{9}{16} - u^2 = -10$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -10-\frac{9}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =-\frac{3}{4} - \frac{13}{4} = -4 s = -\frac{3}{4} + \frac{13}{4} = 2.500$$

Evaluate

$$2\left(2x-5\right)\left(x+4\right)$$