Solve 21x^2-x-2

$$21 { x }^{ 2 } -x-2$$

Factor

$$\left(3x-1\right)\left(7x+2\right)$$

Solution Steps

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

$$a+b=-1 ab=21\left(-2\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=-7 b=6$$

Rewrite 21x^{2}-x-2 as \left(21x^{2}-7x\right)+\left(6x-2\right).

$$\left(21x^{2}-7x\right)+\left(6x-2\right)$$

Factor out 7x in the first and 2 in the second group.

$$7x\left(3x-1\right)+2\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(7x+2\right)$$

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.

$$21x^{2}-x-2=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 21\left(-2\right)}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-\left(-1\right)±\sqrt{1-84\left(-2\right)}}{2\times 21}$$

Multiply -84 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+168}}{2\times 21}$$

Add 1 to 168.

$$x=\frac{-\left(-1\right)±\sqrt{169}}{2\times 21}$$

Take the square root of 169.

$$x=\frac{-\left(-1\right)±13}{2\times 21}$$

The opposite of -1 is 1.

$$x=\frac{1±13}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{1±13}{42}$$

Now solve the equation x=\frac{1±13}{42} when ± is plus. Add 1 to 13.

$$x=\frac{14}{42}$$

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{1±13}{42} when ± is minus. Subtract 13 from 1.

$$x=-\frac{12}{42}$$

Reduce the fraction \frac{-12}{42} to lowest terms by extracting and canceling out 6.

$$x=-\frac{2}{7}$$

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

$$21x^{2}-x-2=21\left(x-\frac{1}{3}\right)\left(x-\left(-\frac{2}{7}\right)\right)$$

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

$$21x^{2}-x-2=21\left(x-\frac{1}{3}\right)\left(x+\frac{2}{7}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}-x-2=21\times \frac{3x-1}{3}\left(x+\frac{2}{7}\right)$$

Add \frac{2}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}-x-2=21\times \frac{3x-1}{3}\times \frac{7x+2}{7}$$

Multiply \frac{3x-1}{3} times \frac{7x+2}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}-x-2=21\times \frac{\left(3x-1\right)\left(7x+2\right)}{3\times 7}$$

Multiply 3 times 7.

$$21x^{2}-x-2=21\times \frac{\left(3x-1\right)\left(7x+2\right)}{21}$$

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

$$21x^{2}-x-2=\left(3x-1\right)\left(7x+2\right)$$

Evaluate

$$\left(3x-1\right)\left(7x+2\right)$$