Solve ((2x+1)(x+2))/(x-3)=0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ (2x+1)(x+2) }{ x-3 } =0$$

Solve for x

$x=-2$
$x=-\frac{1}{2}=-0.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$\left(2x+1\right)\left(x+2\right)=0$$

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

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

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

$$a+b=5$$ $$ab=2\times 2=4$$

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 $4$.

$$1,4$$ $$2,2$$

Calculate the sum for each pair.

$$1+4=5$$ $$2+2=4$$

The solution is the pair that gives sum $5$.

$$a=1$$ $$b=4$$

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

$$\left(2x^{2}+x\right)+\left(4x+2\right)$$

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

$$x\left(2x+1\right)+2\left(2x+1\right)$$

Factor out common term $2x+1$ by using distributive property.

$$\left(2x+1\right)\left(x+2\right)$$

To find equation solutions, solve $2x+1=0$ and $x+2=0$.

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

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$\left(2x+1\right)\left(x+2\right)=0$$

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

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

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

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 2}}{2\times 2}$$

Square $5$.

$$x=\frac{-5±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-5±\sqrt{25-8\times 2}}{2\times 2}$$

Multiply $-8$ times $2$.

$$x=\frac{-5±\sqrt{25-16}}{2\times 2}$$

Add $25$ to $-16$.

$$x=\frac{-5±\sqrt{9}}{2\times 2}$$

Take the square root of $9$.

$$x=\frac{-5±3}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-5±3}{4}$$

Now solve the equation $x=\frac{-5±3}{4}$ when $±$ is plus. Add $-5$ to $3$.

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

Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-5±3}{4}$ when $±$ is minus. Subtract $3$ from $-5$.

$$x=-\frac{8}{4}$$

Divide $-8$ by $4$.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$\left(2x+1\right)\left(x+2\right)=0$$

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

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

Subtract $2$ from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by $2$.

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

Dividing by $2$ undoes the multiplication by $2$.

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

Divide $-2$ by $2$.

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

Divide $\frac{5}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{4}$. Then add the square of $\frac{5}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $\frac{5}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Add $-1$ to $\frac{25}{16}$.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Factor $x^{2}+\frac{5}{2}x+\frac{25}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{3}{4}$$ $$x+\frac{5}{4}=-\frac{3}{4}$$

Subtract $\frac{5}{4}$ from both sides of the equation.

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