Solve 3x^2-x-10

$$3 { x }^{ 2 } -x-10$$

Factor

$$\left(x-2\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-10\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite 3x^{2}-x-10 as \left(3x^{2}-6x\right)+\left(5x-10\right).

$$\left(3x^{2}-6x\right)+\left(5x-10\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(x-2\right)+5\left(x-2\right)$$

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

$$\left(x-2\right)\left(3x+5\right)$$

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.

$$3x^{2}-x-10=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-10\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-10\right)}}{2\times 3}$$

Multiply -12 times -10.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 3}$$

Add 1 to 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-\left(-1\right)±11}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±11}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±11}{6}$$

Now solve the equation x=\frac{1±11}{6} when ± is plus. Add 1 to 11.

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

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{1±11}{6} when ± is minus. Subtract 11 from 1.

$$x=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{3}$$

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{5}{3} for x_{2}.

$$3x^{2}-x-10=3\left(x-2\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$3x^{2}-x-10=3\left(x-2\right)\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-x-10=3\left(x-2\right)\times \frac{3x+5}{3}$$

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

$$3x^{2}-x-10=\left(x-2\right)\left(3x+5\right)$$

Evaluate

$$\left(x-2\right)\left(3x+5\right)$$