1. Expanding "a factor times a sum"

If a factor $a$ is multiplied by the sum $(b+c)$, i.e., $a \cdot (b+c)$, then the distributive law has to be applied to expand this product.

Rule 1:

$a \cdot (b + c) = a \cdot b + a\cdot c$

Examples:

$5a (3b-2c) = 15ab -10ac$

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

$- 10(a-b+c-d) = -10a+10b-10c+10d $

2. Expanding "a sum times a sum"

How can a product of two sums be expanded? The following rule helps to solve this problem.

Rule 2:

For a product of two sums, each sumand of the first sum is multiplied by all summands of the second sum.

Examples:

$(a+b) \cdot(c+d) =ac+ad+bc+bd$

$(2x+y) \cdot(5+3z) =10x+6xz+5y+3yz$

$(-a+b) \cdot(3c+7d) =-3ac-7ad+3bd+7bd$

$(4x-3y) \cdot(-x+6y^2+10xy) =-4x^2+24xy^2+40x^2y+3xy-18y^3-30xy^2$

Two remarks:

• The first two examples are special cases of the Binomial Formulas.

• The result in the last example can be written more compactly as $-4x^2+40x^2y+3xy-18y^3-6xy^2$.

3. Expanding "a sum multiplied by a sum multiplied by another sum and so forth"

How can a product of several sums be expanded? A useful approach is explained by two examples.

Example 1:

\[(a+b) \cdot(c+d) \cdot(e+f)=(a+b) \cdot(ce+cf+de+df)\]\[=ace+acf+ade+adf +bce+bcf+bde+bdf\]

In a first step, the first term $(a+b)$ remains unchanged, while the product $(c+d) \cdot(e+f)$ is expanded. In a second step, the outcome is mutiplied by $(a+b)$. Alternatively, one can expand the first and second term first, while leaving the third term unchanged. Then, the result of the expanded first and second terms is multiplied by the third term. It is easy to see that the final result is identical:

\[{(a+b) \cdot(c+d) \cdot(e+f)=(ac+ad+bc+bd) \cdot (e+f)}\]\[{=ace+ade+bce+bde+acf+adf+bcf+bdf}\]

Example 2:

\[{{\left( a+b \right)}^{3}}=\left[ a+b \right]\cdot \left( a+b \right)\cdot \left( a+b \right)=\left[ a+b \right]\left( {{a}^{2}}+2ab+{{b}^{2}} \right)\]\[={{a}^{3}}+2{{a}^{2}}b+a{{b}^{2}}+{{a}^{2}}b+2a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\]

1)

$6\left( x+y \right)=$

2)

$-4(2a+3b)=$

3)

$(4ab-b+5c)(-2d)=$

4)

$4x\left( -2x \right)\left( -x \right)3x\left( -x \right)=$

5)

$\left( x+2y \right)\left( 3x-y \right)=$

6)

$\left( 3{{x}^{2}}-xy-4{{y}^{2}} \right)\left( -{{x}^{2}}+2xy+{{y}^{2}} \right)=$

7)

$\left( 1-a \right)\left( 5-b \right)\left( 5+b \right)=$

8)

$\left( a-xb \right)\left( c+3d \right)=$

9)

$\left( 5-3a \right)\left( 2-6b \right)=$

10)

$\left( 3a+b \right)\left( 4+3c \right)=$

Solution

A helpful tutorial dealing with expanding one bracket can be found on:

ExamSolutions

A helpful tutorial dealing with expanding two or more brackets can be found on:

ExamSolutions