Solve Complete the factoring . x^ 2 +10x+25=(x+5)( | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

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

a⋅(b-2)=3b

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

a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { x ^ { 2 } + 10 x + 25 = ( x + 5 ) ( \quad ) } \end{array} \right.$$

Solve for x

$x=-5$
$x=-4$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Subtract $x$ from both sides.

$$x^{2}+10x+25-x=5$$

Combine $10x$ and $-x$ to get $9x$.

$$x^{2}+9x+25=5$$

Subtract $5$ from both sides.

$$x^{2}+9x+25-5=0$$

Subtract $5$ from $25$ to get $20$.

$$x^{2}+9x+20=0$$

To solve the equation, factor $x^{2}+9x+20$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=9$$ $$ab=20$$

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 $20$.

$$1,20$$ $$2,10$$ $$4,5$$

Calculate the sum for each pair.

$$1+20=21$$ $$2+10=12$$ $$4+5=9$$

The solution is the pair that gives sum $9$.

$$a=4$$ $$b=5$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x+4\right)\left(x+5\right)$$

To find equation solutions, solve $x+4=0$ and $x+5=0$.

$$x=-4$$ $$x=-5$$

Steps Using Factoring By Grouping

Subtract $x$ from both sides.

$$x^{2}+10x+25-x=5$$

Combine $10x$ and $-x$ to get $9x$.

$$x^{2}+9x+25=5$$

Subtract $5$ from both sides.

$$x^{2}+9x+25-5=0$$

Subtract $5$ from $25$ to get $20$.

$$x^{2}+9x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+20$. To find $a$ and $b$, set up a system to be solved.

$$a+b=9$$ $$ab=1\times 20=20$$

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 $20$.

$$1,20$$ $$2,10$$ $$4,5$$

Calculate the sum for each pair.

$$1+20=21$$ $$2+10=12$$ $$4+5=9$$

The solution is the pair that gives sum $9$.

$$a=4$$ $$b=5$$

Rewrite $x^{2}+9x+20$ as $\left(x^{2}+4x\right)+\left(5x+20\right)$.

$$\left(x^{2}+4x\right)+\left(5x+20\right)$$

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

$$x\left(x+4\right)+5\left(x+4\right)$$

Factor out common term $x+4$ by using distributive property.

$$\left(x+4\right)\left(x+5\right)$$

To find equation solutions, solve $x+4=0$ and $x+5=0$.

$$x=-4$$ $$x=-5$$

Steps Using the Quadratic Formula

Subtract $x$ from both sides.

$$x^{2}+10x+25-x=5$$

Combine $10x$ and $-x$ to get $9x$.

$$x^{2}+9x+25=5$$

Subtract $5$ from both sides.

$$x^{2}+9x+25-5=0$$

Subtract $5$ from $25$ to get $20$.

$$x^{2}+9x+20=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $9$ for $b$, and $20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-9±\sqrt{9^{2}-4\times 20}}{2}$$

Square $9$.

$$x=\frac{-9±\sqrt{81-4\times 20}}{2}$$

Multiply $-4$ times $20$.

$$x=\frac{-9±\sqrt{81-80}}{2}$$

Add $81$ to $-80$.

$$x=\frac{-9±\sqrt{1}}{2}$$

Take the square root of $1$.

$$x=\frac{-9±1}{2}$$

Now solve the equation $x=\frac{-9±1}{2}$ when $±$ is plus. Add $-9$ to $1$.

$$x=-\frac{8}{2}$$

Divide $-8$ by $2$.

$$x=-4$$

Now solve the equation $x=\frac{-9±1}{2}$ when $±$ is minus. Subtract $1$ from $-9$.

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

Divide $-10$ by $2$.

$$x=-5$$

The equation is now solved.

$$x=-4$$ $$x=-5$$

Steps for Completing the Square

Subtract $x$ from both sides.

$$x^{2}+10x+25-x=5$$

Combine $10x$ and $-x$ to get $9x$.

$$x^{2}+9x+25=5$$

Subtract $25$ from both sides.

$$x^{2}+9x=5-25$$

Subtract $25$ from $5$ to get $-20$.

$$x^{2}+9x=-20$$

Divide $9$, the coefficient of the $x$ term, by $2$ to get $\frac{9}{2}$. Then add the square of $\frac{9}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=-20+\left(\frac{9}{2}\right)^{2}$$

Square $\frac{9}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add $-20$ to $\frac{81}{4}$.

$$x^{2}+9x+\frac{81}{4}=\frac{1}{4}$$

Factor $x^{2}+9x+\frac{81}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{1}{2}$$ $$x+\frac{9}{2}=-\frac{1}{2}$$

Subtract $\frac{9}{2}$ from both sides of the equation.

$$x=-4$$ $$x=-5$$