Solve 3x^2-20x-100=

$$3 { x }^{ 2 } -20x-100=$$

Factor

$$\left(x-10\right)\left(3x+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-100. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\left(-100\right)=-300$$

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 -300.

$$1,-300 2,-150 3,-100 4,-75 5,-60 6,-50 10,-30 12,-25 15,-20$$

Calculate the sum for each pair.

$$1-300=-299 2-150=-148 3-100=-97 4-75=-71 5-60=-55 6-50=-44 10-30=-20 12-25=-13 15-20=-5$$

The solution is the pair that gives sum -20.

$$a=-30 b=10$$

Rewrite 3x^{2}-20x-100 as \left(3x^{2}-30x\right)+\left(10x-100\right).

$$\left(3x^{2}-30x\right)+\left(10x-100\right)$$

Factor out 3x in the first and 10 in the second group.

$$3x\left(x-10\right)+10\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(3x+10\right)$$

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.

$$3x^{2}-20x-100=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-100\right)}}{2\times 3}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-100\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-100\right)}}{2\times 3}$$

Multiply -12 times -100.

$$x=\frac{-\left(-20\right)±\sqrt{400+1200}}{2\times 3}$$

Add 400 to 1200.

$$x=\frac{-\left(-20\right)±\sqrt{1600}}{2\times 3}$$

Take the square root of 1600.

$$x=\frac{-\left(-20\right)±40}{2\times 3}$$

The opposite of -20 is 20.

$$x=\frac{20±40}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{20±40}{6}$$

Now solve the equation x=\frac{20±40}{6} when ± is plus. Add 20 to 40.

$$x=\frac{60}{6}$$

Divide 60 by 6.

$$x=10$$

Now solve the equation x=\frac{20±40}{6} when ± is minus. Subtract 40 from 20.

$$x=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -\frac{10}{3} for x_{2}.

$$3x^{2}-20x-100=3\left(x-10\right)\left(x-\left(-\frac{10}{3}\right)\right)$$

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

$$3x^{2}-20x-100=3\left(x-10\right)\left(x+\frac{10}{3}\right)$$

Add \frac{10}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-20x-100=3\left(x-10\right)\times \frac{3x+10}{3}$$

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

$$3x^{2}-20x-100=\left(x-10\right)\left(3x+10\right)$$

Evaluate

$$\left(x-10\right)\left(3x+10\right)$$