# Algebraic expressions

Multiplying out brackets (easy)

## Multiplying out single brackets

To multiply out/expand brackets means multiplying everything inside the bracket by the letter or variable outside the bracket. We call this the Distributive Rule:

\$\$m(a + b) = ma + mb\$\$
Example:
\$\$4(3y+7)=12y+28\$\$
\$\$z(z+y−2x)=z^2+zy−2xz\$\$

## Multiplying out double brackets

Two brackets next to each other means the brackets need to be multiplied together. Every term in the first bracket has to be multiplied by every term in the second bracket:

\$\$(m+n)(a+b)=ma+mb+na+nb\$\$
Example:
\$\$(3y+2)(4−y)=\$\$
\$\$=3y×4−3y×y+2×4−2×y=\$\$
\$\$=12y−3y^2+8−2y=\$\$
\$\$=10y−3y^2+8\$\$

Using formulae (difficult)

\$\$a^2=a×a\$\$

\$\$a^1=a\$\$

\$\$a^0=1\$\$

\$\$a^{−1}=1/a\$\$

\$\$a^{−n}=1/{a^n}\$\$

\$\$a^m×a^n=a^m a^n=a^{(m+n)}\$\$

\$\${a^m}/{a^n}=a^{m−n}\$\$

\$\$(a^m)^n=a^{(m×n)}\$\$

\$\$a^{1/n}=√^{n}a\$\$

\$\$√^n{a}√^n{b} = √^n{ab}\$\$

\$\$√{ab} = √a×√b\$\$

\$\$(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\$\$

\$\$(a−b)^2=(a−b)(a−b)=a^2−2ab+b^2\$\$

\$\$a^2−b^2=(a+b)(a−b)\$\$

\$\$(a+b)^3=a^3+3a^2b+3ab^2+b^3\$\$

\$\$(a−b)^3=a^3−3a^2b+3ab^2−b^3\$\$

Factoring (easy)

## Factorization

Factorization or factoring means writing a number or algebraic expression as a product of several factors, e.g.:

\$\$ax+bx=x(a+b)\$\$

To do the factorization, you need to find the greatest common factor, e.g.:

\$\$2x+6xy−8x^2=2x(1+3y−4x)\$\$

Also:

\$\$7x^2+7x+3x+3=\$\$
\$\$=7x(x+1)+3(x+1)=\$\$
\$\$=(7x+3)(x+1)\$\$

Factoring (difficult)

Factoring/factorizing/factorization is the process of finding the factors. For quadratics (a degree-two polynomials: \$ax^2+bx+c\$) it looks like the following:

\$\$ax^2+bx+c=(ex+f)(gx+h)\$\$

Numbers f and h are called polynomial factors.

### Simple version (a=1)

For \$ax^2+bx+c\$ where \$a=1 (e.g. x^2+7x+6)\$ you have to find 2 numbers (polynomial factors f, h) that are factors of c and if added together you will get b.
In \$x^2+7x+6\$, c is 6. Number 6 has the following factors: 2×3 ; 6×1. Just the sum of the factors 6 and 1 is equal to b (7). So our f and h are 6 and 1.

\$\$x^2+7x+6=(x+6)(x+1)\$\$

Let's prove it:

\$\$(x+6)(x+1)=x^2+x+6x+6=\$\$
\$\$=x^2+7x+6\$\$

The situation is different with some of the terms being negative:

\$\$x^2−6x+8\$\$

8 is positive, so the factors will always be either both positive (+) or both negative (−): 2×4 ; (−2)×(−4) ; 1×8 ; (−1)×(−8)

If the middle coefficient is negative (−6) we have to use negative factors. (−6)=(−2)+(−4). So our f and h are: −2 and −4:

\$\$x^2−6x+8=(x−2)(x−4)\$\$

Let's prove it:

\$\$(x−2)(x−4)=x^2−4x−2x+8=\$\$
\$\$x^2−6x+8\$\$

### Difficult version (a≠1)

For \$ax^2+bx+c\$ where a≠1 (e.g. \$2x^2+13x+6\$) you have to find 2 numbers that are factors of a×c and add to give you b.

\$\$2x^2+13x+6\$\$
\$\$a×c=2×6=12\$\$

Let's find factors of 12: 2×6 ; 3×4 ; 12×1

Next, find the pair of factors that adds to b (13), it's only the last one: b=13=12+1 .
Now, split the middle and finish by grouping (sometimes we need to take two tries to find which of the two middle addends should go first):

\$\$2x^2+12x+x+6=\$\$
\$\$=2x(x+6)+(x+6)=\$\$
\$\$=(x+6)(2x+1)\$\$

Let's prove it:

\$\$(x+6)(2x+1)=2x^2+x+12x+6\$\$

Another example with negative terms:

\$\$3x^2+10x−8\$\$
\$\$a×c=3×(−8)=(−24)\$\$

Let's find factors of (−24): (−24)×1 ; (−1)×24 ; (−3)×8 ; 8×(−3) ; 12×(−2) ; ...
We have the pair that adds up to b (10): 10=12+(−2)
Split the middle term:

\$\$3x^2+10x−8=\$\$
\$\$=3x^2+12x−2x−8=\$\$
\$\$=3x(x+4)−2(x+4)=(3x−2)(x+4)\$\$

Let's prove it:

\$\$(3x−2)(x+4)=3x^2+12x−2x−8=\$\$
\$\$=3x^2+10x−8\$\$

Algebraic fractions (easy)

## Algebraic fractions

To simplify algebraic fractions, you need to keep basic algebraic formulae and stick to order of operations:

Step 1:
Evaluate the parentheses
Step 2:
Evaluate the powers (exponents and roots)
Step 3:
Multiply or divide from left to right
Step 4:

\$\$a^2=a×a\$\$

\$\$a^1=a\$\$

\$\$a^0=1\$\$

\$\$a^{−1}=1/a\$\$

\$\$a^{−n}=1/{a^n}\$\$

\$\$a^m×a^n=a^m a^n=a^{(m+n)}\$\$

\$\${a^m}/{a^n}=a^{m−n}\$\$

\$\$(a^m)^n=a^{(m×n)}\$\$

\$\$a^{1/n}=√^{n}a\$\$

\$\$√^n{a}√^n{b} = √^n{ab}\$\$

\$\$√{ab} = √a×√b\$\$

\$\$(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\$\$

\$\$(a−b)^2=(a−b)(a−b)=a^2−2ab+b^2\$\$

\$\$a^2−b^2=(a+b)(a−b)\$\$

\$\$(a+b)^3=a^3+3a^2b+3ab^2+b^3\$\$

\$\$(a−b)^3=a^3−3a^2b+3ab^2−b^3\$\$