Solve (1+2x)(3-x)+x^2=9

$$( 1 + 2 x ) ( 3 - x ) + x ^ { 2 } = 9$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 1+2x by 3-x and combine like terms.

$$3+5x-2x^{2}+x^{2}=9$$

Combine -2x^{2} and x^{2} to get -x^{2}.

$$3+5x-x^{2}=9$$

Subtract 9 from both sides.

$$3+5x-x^{2}-9=0$$

Subtract 9 from 3 to get -6.

$$-6+5x-x^{2}=0$$

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

$$-x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -x^{2}+5x-6 as \left(-x^{2}+3x\right)+\left(2x-6\right).

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

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

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

Factor out common term x-3 by using distributive property.

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

To find equation solutions, solve x-3=0 and -x+2=0.

$$x=3 x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1+2x by 3-x and combine like terms.

$$3+5x-2x^{2}+x^{2}=9$$

Combine -2x^{2} and x^{2} to get -x^{2}.

$$3+5x-x^{2}=9$$

Subtract 9 from both sides.

$$3+5x-x^{2}-9=0$$

Subtract 9 from 3 to get -6.

$$-6+5x-x^{2}=0$$

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.

$$-x^{2}+5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-5±\sqrt{25-24}}{2\left(-1\right)}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-5±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±1}{-2}$$

Now solve the equation x=\frac{-5±1}{-2} when ± is plus. Add -5 to 1.

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

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-5±1}{-2} when ± is minus. Subtract 1 from -5.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Steps for Completing the Square

Use the distributive property to multiply 1+2x by 3-x and combine like terms.

$$3+5x-2x^{2}+x^{2}=9$$

Combine -2x^{2} and x^{2} to get -x^{2}.

$$3+5x-x^{2}=9$$

Subtract 3 from both sides.

$$5x-x^{2}=9-3$$

Subtract 3 from 9 to get 6.

$$5x-x^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+5x=6$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=\frac{6}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}-5x=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$