Solve 20x^2+11x-3

$$20 { x }^{ 2 } +11x-3$$

Factor

$$\left(5x-1\right)\left(4x+3\right)$$

Solution Steps

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

$$a+b=11 ab=20\left(-3\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=-4 b=15$$

Rewrite 20x^{2}+11x-3 as \left(20x^{2}-4x\right)+\left(15x-3\right).

$$\left(20x^{2}-4x\right)+\left(15x-3\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(5x-1\right)+3\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(4x+3\right)$$

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.

$$20x^{2}+11x-3=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 20\left(-3\right)}}{2\times 20}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 20\left(-3\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-11±\sqrt{121-80\left(-3\right)}}{2\times 20}$$

Multiply -80 times -3.

$$x=\frac{-11±\sqrt{121+240}}{2\times 20}$$

Add 121 to 240.

$$x=\frac{-11±\sqrt{361}}{2\times 20}$$

Take the square root of 361.

$$x=\frac{-11±19}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-11±19}{40}$$

Now solve the equation x=\frac{-11±19}{40} when ± is plus. Add -11 to 19.

$$x=\frac{8}{40}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-11±19}{40} when ± is minus. Subtract 19 from -11.

$$x=-\frac{30}{40}$$

Reduce the fraction \frac{-30}{40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{5} for x_{1} and -\frac{3}{4} for x_{2}.

$$20x^{2}+11x-3=20\left(x-\frac{1}{5}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$20x^{2}+11x-3=20\left(x-\frac{1}{5}\right)\left(x+\frac{3}{4}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+11x-3=20\times \frac{5x-1}{5}\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+11x-3=20\times \frac{5x-1}{5}\times \frac{4x+3}{4}$$

Multiply \frac{5x-1}{5} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+11x-3=20\times \frac{\left(5x-1\right)\left(4x+3\right)}{5\times 4}$$

Multiply 5 times 4.

$$20x^{2}+11x-3=20\times \frac{\left(5x-1\right)\left(4x+3\right)}{20}$$

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

$$20x^{2}+11x-3=\left(5x-1\right)\left(4x+3\right)$$

Evaluate

$$\left(5x-1\right)\left(4x+3\right)$$