Solve 21x^2+13x-20

$$21 { x }^{ 2 } +13x-20$$

Factor

$$\left(7x-5\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=13 ab=21\left(-20\right)=-420$$

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 -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 13.

$$a=-15 b=28$$

Rewrite 21x^{2}+13x-20 as \left(21x^{2}-15x\right)+\left(28x-20\right).

$$\left(21x^{2}-15x\right)+\left(28x-20\right)$$

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

$$3x\left(7x-5\right)+4\left(7x-5\right)$$

Factor out common term 7x-5 by using distributive property.

$$\left(7x-5\right)\left(3x+4\right)$$

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.

$$21x^{2}+13x-20=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 21\left(-20\right)}}{2\times 21}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 21\left(-20\right)}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-13±\sqrt{169-84\left(-20\right)}}{2\times 21}$$

Multiply -84 times -20.

$$x=\frac{-13±\sqrt{169+1680}}{2\times 21}$$

Add 169 to 1680.

$$x=\frac{-13±\sqrt{1849}}{2\times 21}$$

Take the square root of 1849.

$$x=\frac{-13±43}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-13±43}{42}$$

Now solve the equation x=\frac{-13±43}{42} when ± is plus. Add -13 to 43.

$$x=\frac{30}{42}$$

Reduce the fraction \frac{30}{42} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{7}$$

Now solve the equation x=\frac{-13±43}{42} when ± is minus. Subtract 43 from -13.

$$x=-\frac{56}{42}$$

Reduce the fraction \frac{-56}{42} to lowest terms by extracting and canceling out 14.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{7} for x_{1} and -\frac{4}{3} for x_{2}.

$$21x^{2}+13x-20=21\left(x-\frac{5}{7}\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$21x^{2}+13x-20=21\left(x-\frac{5}{7}\right)\left(x+\frac{4}{3}\right)$$

Subtract \frac{5}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+13x-20=21\times \frac{7x-5}{7}\left(x+\frac{4}{3}\right)$$

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

$$21x^{2}+13x-20=21\times \frac{7x-5}{7}\times \frac{3x+4}{3}$$

Multiply \frac{7x-5}{7} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+13x-20=21\times \frac{\left(7x-5\right)\left(3x+4\right)}{7\times 3}$$

Multiply 7 times 3.

$$21x^{2}+13x-20=21\times \frac{\left(7x-5\right)\left(3x+4\right)}{21}$$

Cancel out 21, the greatest common factor in 21 and 21.

$$21x^{2}+13x-20=\left(7x-5\right)\left(3x+4\right)$$

Evaluate

$$\left(7x-5\right)\left(3x+4\right)$$