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

### Solved Examples

**Question 1:**Solve the equation, 5(x - 3)

^{2}- 20 = 0

**Solution:**

Given 5(x - 3)

Rewrite equation with square on one side

=> 5(x - 3)

=> 5(x

=> 5(x

=> 5x

=> 5x

=> 5x

Solve for x,

=> 5x

=> 5x

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

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

either 5x - 5 = 0 or x - 5 = 0

=> 5x = 5 or x = 5

=> x = 1 or x = 5

^{2}- 20 = 0**Step 1:**Rewrite equation with square on one side

=> 5(x - 3)

^{2}- 20 = 0=> 5(x

^{2}+ 3^{2}- 2 * x * 3) - 20 = 0=> 5(x

^{2}+ 9 - 6x) - 20 = 0=> 5x

^{2}+ 5 * 9 - 5 * 6x - 20 = 0=> 5x

^{2}+ 45 - 30x - 20 = 0=> 5x

^{2}- 30x + 25 = 0**Step 2:**Solve for x,

=> 5x

^{2}- 30x + 25 = 0=> 5x

^{2}- 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

put x - 3 = 0

=> x = 3

=> 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

**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

Given Perimeter of the rectangle = 38

Perimeter of the rectangle = 2(Length + Width)

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

=> 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

Add 2 to both sides

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

=> 21 = 3x

divide each side by 3

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

=> 7 = x

or x = 7

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.

Therefore width of rectangle = 2x - 2

Step 1: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

Add 2 to both sides

=> 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**