Solve 20x^2+39x-11

$$20 x ^ { 2 } + 39 x - 11$$

Factor

$$\left(4x-1\right)\left(5x+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-11. To find a and b, set up a system to be solved.

$$a+b=39 ab=20\left(-11\right)=-220$$

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 -220.

$$-1,220 -2,110 -4,55 -5,44 -10,22 -11,20$$

Calculate the sum for each pair.

$$-1+220=219 -2+110=108 -4+55=51 -5+44=39 -10+22=12 -11+20=9$$

The solution is the pair that gives sum 39.

$$a=-5 b=44$$

Rewrite 20x^{2}+39x-11 as \left(20x^{2}-5x\right)+\left(44x-11\right).

$$\left(20x^{2}-5x\right)+\left(44x-11\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(4x-1\right)+11\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(5x+11\right)$$

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.

$$20x^{2}+39x-11=0$$

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.

$$x=\frac{-39±\sqrt{39^{2}-4\times 20\left(-11\right)}}{2\times 20}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\times 20\left(-11\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-39±\sqrt{1521-80\left(-11\right)}}{2\times 20}$$

Multiply -80 times -11.

$$x=\frac{-39±\sqrt{1521+880}}{2\times 20}$$

Add 1521 to 880.

$$x=\frac{-39±\sqrt{2401}}{2\times 20}$$

Take the square root of 2401.

$$x=\frac{-39±49}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-39±49}{40}$$

Now solve the equation x=\frac{-39±49}{40} when ± is plus. Add -39 to 49.

$$x=\frac{10}{40}$$

Reduce the fraction \frac{10}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-39±49}{40} when ± is minus. Subtract 49 from -39.

$$x=-\frac{88}{40}$$

Reduce the fraction \frac{-88}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{11}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and -\frac{11}{5} for x_{2}.

$$20x^{2}+39x-11=20\left(x-\frac{1}{4}\right)\left(x-\left(-\frac{11}{5}\right)\right)$$

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

$$20x^{2}+39x-11=20\left(x-\frac{1}{4}\right)\left(x+\frac{11}{5}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+39x-11=20\times \frac{4x-1}{4}\left(x+\frac{11}{5}\right)$$

Add \frac{11}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+39x-11=20\times \frac{4x-1}{4}\times \frac{5x+11}{5}$$

Multiply \frac{4x-1}{4} times \frac{5x+11}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+39x-11=20\times \frac{\left(4x-1\right)\left(5x+11\right)}{4\times 5}$$

Multiply 4 times 5.

$$20x^{2}+39x-11=20\times \frac{\left(4x-1\right)\left(5x+11\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20x^{2}+39x-11=\left(4x-1\right)\left(5x+11\right)$$

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 20

$$x ^ 2 +\frac{39}{20}x -\frac{11}{20} = 0$$

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

$$r + s = -\frac{39}{20} rs = -\frac{11}{20}$$

Two numbers r and s sum up to -\frac{39}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{39}{20} = -\frac{39}{40}. 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>

$$r = -\frac{39}{40} - u s = -\frac{39}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{11}{20}

$$(-\frac{39}{40} - u) (-\frac{39}{40} + u) = -\frac{11}{20}$$

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

$$\frac{1521}{1600} - u^2 = -\frac{11}{20}$$

Simplify the expression by subtracting \frac{1521}{1600} on both sides

$$-u^2 = -\frac{11}{20}-\frac{1521}{1600} = -\frac{2401}{1600}$$

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

$$u^2 = \frac{2401}{1600} u = \pm\sqrt{\frac{2401}{1600}} = \pm \frac{49}{40}$$

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

$$r =-\frac{39}{40} - \frac{49}{40} = -2.200 s = -\frac{39}{40} + \frac{49}{40} = 0.250$$

Evaluate

$$\left(4x-1\right)\left(5x+11\right)$$