Solve 30+3x-6x^2

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

Factor

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

Solution Steps

Factor out 3.

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

Consider 10+x-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.

$$a+b=1 ab=-2\times 10=-20$$

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 -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.

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

Square 3.

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

Multiply -4 times -6.

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

Multiply 24 times 30.

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

Add 9 to 720.

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

Take the square root of 729.

$$x=\frac{-3±27}{2\left(-6\right)}$$

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.

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

Divide 24 by -12.

$$x=-2$$

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

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

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

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

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}{2} for x_{2}.

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

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

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

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

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

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

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

Evaluate

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