Solve 45+4x-x^2

$$45 + 4 x - x ^ { 2 }$$

Factor

$$-\left(x-9\right)\left(x+5\right)$$

Solution Steps

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

$$-x^{2}+4x+45$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=4 ab=-45=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=9 b=-5$$

Rewrite -x^{2}+4x+45 as \left(-x^{2}+9x\right)+\left(-5x+45\right).

$$\left(-x^{2}+9x\right)+\left(-5x+45\right)$$

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

$$-x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-5\right)$$

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.

$$-x^{2}+4x+45=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 45}}{2\left(-1\right)}$$

Multiply 4 times 45.

$$x=\frac{-4±\sqrt{16+180}}{2\left(-1\right)}$$

Add 16 to 180.

$$x=\frac{-4±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$x=\frac{-4±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±14}{-2}$$

Now solve the equation x=\frac{-4±14}{-2} when ± is plus. Add -4 to 14.

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

Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{-4±14}{-2} when ± is minus. Subtract 14 from -4.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 9 for x_{2}.

$$-x^{2}+4x+45=-\left(x-\left(-5\right)\right)\left(x-9\right)$$

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

$$-x^{2}+4x+45=-\left(x+5\right)\left(x-9\right)$$

Evaluate

$$-\left(x-9\right)\left(x+5\right)$$