# Algebra Expression Solver

Algebraic expressions are the backbone of the subject. Staring from simple linear equations, we move on to polynomials and quadratic equations. Equations use numerals, variables, integers and exponents to express generalizations in the form of equations. Complicated as they may seem initially, these elaborate equations are quite easy to solve if you know how and which order to break them down in. The few basic rules that algebra follows are all you need to know to get cracking. Online algebra expression solvers can handle any kind of equations and solve them instantly so that you can use them to cross check your work.

## Solving Algebraic Expressions

Algebraic expressions solver available online gives students a step by step guide to solving algebraic equations. These free and easy to use online helpers teach students how to go about expanding an equation so that it becomes simple and easier to solve. Algebra expression solver will help you learn whether addition or division comes first and which of the parentheses gets priority. Easy to use with large graphics and fun online tools, algebra expression solvers make the subject fun and interesting. Once you are familiar with the subject, algebra equation solver also gives great tips and tricks that you can use to save time and solve equations easily.

### Solved Examples

Question 1: Solve the equation, 5(x - 3)2 - 20 = 0
Solution:
Given 5(x - 3)2 - 20 = 0

Step 1:
Rewrite equation with square on one side

=> 5(x - 3)2 - 20 = 0

=> 5(x2 + 32 - 2 * x * 3) - 20 = 0

=> 5(x2 + 9 - 6x) - 20 = 0

=> 5x2 + 5 * 9 - 5 * 6x - 20 = 0

=> 5x2 + 45 - 30x - 20 = 0

=> 5x2 - 30x + 25 = 0

Step 2:
Solve for x,

=> 5x2 - 30x + 25 = 0

=> 5x2 - 25x - 5x + 25 = 0

=> 5x(x - 5) - 5(x - 5) = 0

=> (5x - 5)(x - 5) = 0

Step 3:
either 5x - 5 = 0    or   x - 5 = 0

=> 5x = 5     or     x = 5

=> x = 1       or     x = 5

The solutions of the given equation are 1 and 5.

Question 2: Show that x - 3 is a factor of 4$x^3$ - 6$x^2$ + 12x + 10.
Solution:
Let f(x) = 4$x^3$ - 6$x^2$ + 12x + 10

Step 1:
To check x - 3 is a factor of f(x),

put x - 3 = 0

=> x = 3

Step 2:
x - 3 is a factor of f(x) if f(3) = 0

=> f(3) = 4($3^3$) - 6($3^2$) + 12 * 3 + 10

=> f(3) = 4 * 27 - 6 * 9 + 36 + 10

=> f(3) = 108 - 54 + 46

=> f(3) = 100

=> f(3) $\neq$ 0

=> x - 3 is not a factor of f(x). answer

Question 3: The perimeter of a rectangle is 38 in. the width of the rectangle is 2 in less than twice the length. Find the dimensions of the rectangle.
Solution:
Let the length of the rectangle = x
Therefore width of rectangle = 2x - 2

Step 1:

Given Perimeter of the rectangle = 38

Perimeter of the rectangle = 2(Length + Width)

=> 38 = 2(x + (2x - 2))

Step 2:
Solve for x,

=> 38 = 2(x + 2x - 2)

=> 38 = 2(3x - 2)

Divide both sides by 2

=> $\frac{38}{2} = \frac{ 2(3x - 2)}{2}$

=> 19 = 3x - 2

=> 19 + 2 = 3x - 2 + 2

=> 21 = 3x

divide each side by 3

=> $\frac{21}{3} = \frac{3x}{3}$

=> 7 = x

or x = 7

step 3:

So width of the rectangle = 2x - 2 = 2 * 7 - 2

= 14 - 2

= 12

Hence the dimensions of rectangles are: Length = 7 in. and Width = 12 in. answer