Solve 3x^2+x-30

$$3 { x }^{ 2 } +x-30$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-30\right)=-90$$

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 -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=-9 b=10$$

Rewrite 3x^{2}+x-30 as \left(3x^{2}-9x\right)+\left(10x-30\right).

$$\left(3x^{2}-9x\right)+\left(10x-30\right)$$

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

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

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

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

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.

$$3x^{2}+x-30=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-30\right)}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-30\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-30\right)}}{2\times 3}$$

Multiply -12 times -30.

$$x=\frac{-1±\sqrt{1+360}}{2\times 3}$$

Add 1 to 360.

$$x=\frac{-1±\sqrt{361}}{2\times 3}$$

Take the square root of 361.

$$x=\frac{-1±19}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±19}{6}$$

Now solve the equation x=\frac{-1±19}{6} when ± is plus. Add -1 to 19.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{-1±19}{6} when ± is minus. Subtract 19 from -1.

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

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 3 for x_{1} and -\frac{10}{3} for x_{2}.

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

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

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

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

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

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

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

Evaluate

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