Solve -3x-30+6x^2

$$- 3 x - 30 + 6 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out 3.

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

Consider -x-10+2x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=-1 ab=2\left(-10\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite 2x^{2}-x-10 as \left(2x^{2}-5x\right)+\left(4x-10\right).

$$\left(2x^{2}-5x\right)+\left(4x-10\right)$$

Factor out x in the first and 2 in the second group.

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

Factor out common term 2x-5 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}-3x-30=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6\left(-30\right)}}{2\times 6}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6\left(-30\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-24\left(-30\right)}}{2\times 6}$$

Multiply -24 times -30.

$$x=\frac{-\left(-3\right)±\sqrt{9+720}}{2\times 6}$$

Add 9 to 720.

$$x=\frac{-\left(-3\right)±\sqrt{729}}{2\times 6}$$

Take the square root of 729.

$$x=\frac{-\left(-3\right)±27}{2\times 6}$$

The opposite of -3 is 3.

$$x=\frac{3±27}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{3±27}{12}$$

Now solve the equation x=\frac{3±27}{12} when ± is plus. Add 3 to 27.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{3±27}{12} when ± is minus. Subtract 27 from 3.

$$x=-\frac{24}{12}$$

Divide -24 by 12.

$$x=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -2 for x_{2}.

$$6x^{2}-3x-30=6\left(x-\frac{5}{2}\right)\left(x-\left(-2\right)\right)$$

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

$$6x^{2}-3x-30=6\left(x-\frac{5}{2}\right)\left(x+2\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-3x-30=6\times \frac{2x-5}{2}\left(x+2\right)$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}-3x-30=3\left(2x-5\right)\left(x+2\right)$$

Evaluate

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