Solve 4x^2+2x-20

$$4 x ^ { 2 } + 2 x - 20$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 2x^{2}+x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-10\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 2x^{2}+x-10 as \left(2x^{2}-4x\right)+\left(5x-10\right).

$$\left(2x^{2}-4x\right)+\left(5x-10\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$4x^{2}+2x-20=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\left(-20\right)}}{2\times 4}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 4\left(-20\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\left(-20\right)}}{2\times 4}$$

Multiply -16 times -20.

$$x=\frac{-2±\sqrt{4+320}}{2\times 4}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\times 4}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±18}{8}$$

Now solve the equation x=\frac{-2±18}{8} when ± is plus. Add -2 to 18.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{-2±18}{8} when ± is minus. Subtract 18 from -2.

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

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

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

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

$$4x^{2}+2x-20=4\left(x-2\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$4x^{2}+2x-20=4\left(x-2\right)\left(x+\frac{5}{2}\right)$$

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

$$4x^{2}+2x-20=4\left(x-2\right)\times \frac{2x+5}{2}$$

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

$$4x^{2}+2x-20=2\left(x-2\right)\left(2x+5\right)$$

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 4

$$x ^ 2 +\frac{1}{2}x -5 = 0$$

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

$$r + s = -\frac{1}{2} rs = -5$$

Two numbers r and s sum up to -\frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{2} = -\frac{1}{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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -5$$

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

$$\frac{1}{16} - u^2 = -5$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -5-\frac{1}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{1}{4} - \frac{9}{4} = -2.500 s = -\frac{1}{4} + \frac{9}{4} = 2$$

Evaluate

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