One way to expand binomials is by using grids. To do this, assign the terms of the first bracket to the boxes of the grid vertically and assign the terms of the second bracket to the boxes of the grid horizontally. Fill out the grid by multiplying the terms assigned to each box and take the sum of all the products. This will give the expansion of the two brackets.

Another way to expand binomials is to use the method of FOIL to determine the order of expansion. FOIL stands for first, outer, inner and last and helps remember the terms that must be multiplied together to find an expansion of two brackets. FOIL method suggests to multiply the two first terms of the brackets, the ones on the outer side of the brackets, the ones on the inner side of the brackets and the two last terms of the brackets. Taking the sum of the products gives the expansion of the two brackets.

Factorising is the reverse of expanding. A quadratic expression, x² + ax + b, can be factorised to be written as (x + c)(x + d), where the sum of c and d is equal to a and the product of c and d is equal to b. A special case of factorisation, called the difference between the two squares, is given as: x² − y² = (x − y)(x + y).

Algebraic rational functions can be written as a sum or difference of fractions; this is called partial fractions. To do this, the general forms given in the image can be used. The new fractions can then be added by finding common denominators and coefficients of respective powers of variables equated to get simultaneous equations. An example of splitting a fraction into partial fractions is provided below.

(5x + 7)/(x + 3)(x + 2) = A/(x + 3) + B/(x + 2)
multiplying by (x + 3)(x + 2): 5x + 7 = A(x + 2) + B(x + 3)
expanding: 5x + 7 = Ax + 2A + Bx + 3B
collecting like terms: 5x + 7 = x(A + B) + 2A + 3B
equating coefficients of x: 5 = A + B
equating constants: 7 = 2A + 3B
solving simultaneously: A = 8 and B = −3
(5x + 7)/(x + 3) (x + 2) = 8/(x + 3) − 3(x + 2)

The terms of the brackets are written on the outsides of the grid.
The products of the terms are found and added together.
Thus, (14 − x)(8x + 9) = −8x² + 103x + 126.

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(10x² + 17x + 42)/(x + 10)(x² + 9) = A/(x + 10) + (Bx + C)/(x² + 9)
Multiplying both sides by (x + 10)(x² + 9) gives: 10x² + 17x + 42 = A(x² + 9) + (Bx + 9)(x + 10).
Expanding gives: 10x² + 17x + 42 = Ax² + 9A + Bx² + 10Bx + Cx + 10C.
Equating coefficients of x²: 10 = A + B.
Equating coefficients of x: 17 = 10B + C.
Equating constants: 42 = 9A + 10C.
Multiplying the first equation by 9: 90 = 9A + 9B.
Subtracting the third equation from this gives: 48 = 9B − 10C.
Multiplying the second equation by 10 gives: 170 = 100B + 10C.
Adding this to the previous equation gives: 218 = 109B, so B = 2.
Substituting this in the first equation gives: 10 = A + 2, so A = 8.
Substituting B = 2 in the second equation gives 17 = 20 + C, so C = −3.
Therefore, (10x² + 17x + 42)/(x + 10)(x² + 9) = 8/(x + 10) + (2x − 3)/(x² + 9).

8/(x + 10) + (2x − 3)/(x² + 9)

The terms of the brackets are written on the outsides of the grid.
The products of the terms are found and added together.
Then (55 − x²)(55 − x²) = 3025 − 110x² + x⁴.

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first: 6a³ × 2a = 12a⁴
outer: 6a³ × (−b) = −6a³b
inner: −4b × 2a = −8ab
last: −4b × (−b) = 4b²
(6a³ − 4b)(2a − b) = 12a⁴ − 6a³b − 8ab + 4b².

12a⁴ − 6a³b − 8ab + 4b²

The terms of the brackets are written on the outsides of the grid.
The products of the terms are found and added together.
Therefore, (2a − 4ab)(a² + 4) = −4a³b + 2a³ − 16ab + 8a.

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