Solve 3x^2-20x-7

$$3 { x }^{ 2 } -20x-7$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\left(-7\right)=-21$$

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 -21.

$$1,-21 3,-7$$

Calculate the sum for each pair.

$$1-21=-20 3-7=-4$$

The solution is the pair that gives sum -20.

$$a=-21 b=1$$

Rewrite 3x^{2}-20x-7 as \left(3x^{2}-21x\right)+\left(x-7\right).

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

Factor out 3x in 3x^{2}-21x.

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

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

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

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.

$$3x^{2}-20x-7=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-7\right)}}{2\times 3}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-7\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-7\right)}}{2\times 3}$$

Multiply -12 times -7.

$$x=\frac{-\left(-20\right)±\sqrt{400+84}}{2\times 3}$$

Add 400 to 84.

$$x=\frac{-\left(-20\right)±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$x=\frac{-\left(-20\right)±22}{2\times 3}$$

The opposite of -20 is 20.

$$x=\frac{20±22}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{20±22}{6}$$

Now solve the equation x=\frac{20±22}{6} when ± is plus. Add 20 to 22.

$$x=\frac{42}{6}$$

Divide 42 by 6.

$$x=7$$

Now solve the equation x=\frac{20±22}{6} when ± is minus. Subtract 22 from 20.

$$x=-\frac{2}{6}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

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

$$3x^{2}-20x-7=3\left(x-7\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$3x^{2}-20x-7=3\left(x-7\right)\left(x+\frac{1}{3}\right)$$

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

$$3x^{2}-20x-7=3\left(x-7\right)\times \frac{3x+1}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-20x-7=\left(x-7\right)\left(3x+1\right)$$

Evaluate

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