Solve 3x^2+14x-40

$$3 { x }^{ 2 } +14x-40$$

Factor

$$\left(x-2\right)\left(3x+20\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=14 ab=3\left(-40\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 14.

$$a=-6 b=20$$

Rewrite 3x^{2}+14x-40 as \left(3x^{2}-6x\right)+\left(20x-40\right).

$$\left(3x^{2}-6x\right)+\left(20x-40\right)$$

Factor out 3x in the first and 20 in the second group.

$$3x\left(x-2\right)+20\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(3x+20\right)$$

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.

$$3x^{2}+14x-40=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 3\left(-40\right)}}{2\times 3}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 3\left(-40\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-14±\sqrt{196-12\left(-40\right)}}{2\times 3}$$

Multiply -12 times -40.

$$x=\frac{-14±\sqrt{196+480}}{2\times 3}$$

Add 196 to 480.

$$x=\frac{-14±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$x=\frac{-14±26}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-14±26}{6}$$

Now solve the equation x=\frac{-14±26}{6} when ± is plus. Add -14 to 26.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{-14±26}{6} when ± is minus. Subtract 26 from -14.

$$x=-\frac{40}{6}$$

Reduce the fraction \frac{-40}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{20}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -\frac{20}{3} for x_{2}.

$$3x^{2}+14x-40=3\left(x-2\right)\left(x-\left(-\frac{20}{3}\right)\right)$$

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

$$3x^{2}+14x-40=3\left(x-2\right)\left(x+\frac{20}{3}\right)$$

Add \frac{20}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+14x-40=3\left(x-2\right)\times \frac{3x+20}{3}$$

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

$$3x^{2}+14x-40=\left(x-2\right)\left(3x+20\right)$$

Evaluate

$$\left(x-2\right)\left(3x+20\right)$$