Solve 10+3x-x^2

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=3 ab=-10=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=5 b=-2$$

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

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

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

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

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

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

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.

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

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.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\times 10}}{2\left(-1\right)}$$

Multiply 4 times 10.

$$x=\frac{-3±\sqrt{9+40}}{2\left(-1\right)}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±7}{-2}$$

Now solve the equation x=\frac{-3±7}{-2} when ± is plus. Add -3 to 7.

$$x=\frac{4}{-2}$$

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-3±7}{-2} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

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 5 for x_{2}.

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

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

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

Evaluate

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