Solve 20+46x-10x^2

$$20 + 46 x - 10 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(10+23x-5x^{2}\right)$$

Consider 10+23x-5x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}+23x+10$$

Factor the expression by grouping. First, the expression needs to be rewritten as -5x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=23 ab=-5\times 10=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$a=25 b=-2$$

Rewrite -5x^{2}+23x+10 as \left(-5x^{2}+25x\right)+\left(-2x+10\right).

$$\left(-5x^{2}+25x\right)+\left(-2x+10\right)$$

Factor out 5x in the first and 2 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$-10x^{2}+46x+20=0$$

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.

$$x=\frac{-46±\sqrt{46^{2}-4\left(-10\right)\times 20}}{2\left(-10\right)}$$

Square 46.

$$x=\frac{-46±\sqrt{2116-4\left(-10\right)\times 20}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-46±\sqrt{2116+40\times 20}}{2\left(-10\right)}$$

Multiply 40 times 20.

$$x=\frac{-46±\sqrt{2116+800}}{2\left(-10\right)}$$

Add 2116 to 800.

$$x=\frac{-46±\sqrt{2916}}{2\left(-10\right)}$$

Take the square root of 2916.

$$x=\frac{-46±54}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-46±54}{-20}$$

Now solve the equation x=\frac{-46±54}{-20} when ± is plus. Add -46 to 54.

$$x=\frac{8}{-20}$$

Reduce the fraction \frac{8}{-20} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-46±54}{-20} when ± is minus. Subtract 54 from -46.

$$x=-\frac{100}{-20}$$

Divide -100 by -20.

$$x=5$$

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

$$-10x^{2}+46x+20=-10\left(x-\left(-\frac{2}{5}\right)\right)\left(x-5\right)$$

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

$$-10x^{2}+46x+20=-10\left(x+\frac{2}{5}\right)\left(x-5\right)$$

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

$$-10x^{2}+46x+20=-10\times \frac{-5x-2}{-5}\left(x-5\right)$$

Cancel out 5, the greatest common factor in -10 and 5.

$$-10x^{2}+46x+20=2\left(-5x-2\right)\left(x-5\right)$$

Evaluate

$$20+46x-10x^{2}$$