Solve 10-x^2-9x

$$10 - x ^ { 2 } - 9 x$$

Factor

$$\left(1-x\right)\left(x+10\right)$$

Solution Steps

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

$$-x^{2}-9x+10$$

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.

$$a+b=-9 ab=-10=-10$$

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 -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 -9.

$$a=1 b=-10$$

Rewrite -x^{2}-9x+10 as \left(-x^{2}+x\right)+\left(-10x+10\right).

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

Factor out x in the first and 10 in the second group.

$$x\left(-x+1\right)+10\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+10\right)$$

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.

$$-x^{2}-9x+10=0$$

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.

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

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+4\times 10}}{2\left(-1\right)}$$

Multiply 4 times 10.

$$x=\frac{-\left(-9\right)±\sqrt{81+40}}{2\left(-1\right)}$$

Add 81 to 40.

$$x=\frac{-\left(-9\right)±\sqrt{121}}{2\left(-1\right)}$$

Take the square root of 121.

$$x=\frac{-\left(-9\right)±11}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±11}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{9±11}{-2}$$

Now solve the equation x=\frac{9±11}{-2} when ± is plus. Add 9 to 11.

$$x=\frac{20}{-2}$$

Divide 20 by -2.

$$x=-10$$

Now solve the equation x=\frac{9±11}{-2} when ± is minus. Subtract 11 from 9.

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

Divide -2 by -2.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -10 for x_{1} and 1 for x_{2}.

$$-x^{2}-9x+10=-\left(x-\left(-10\right)\right)\left(x-1\right)$$

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

$$-x^{2}-9x+10=-\left(x+10\right)\left(x-1\right)$$

Evaluate

$$\left(1-x\right)\left(x+10\right)$$