Substitute into common formulae
Algebraic fractions
Distributive law I
Distributive law (non-linear terms) II
Substitute into algebraic expressions
Multiply algebraic terms with indices
Divide algebraic terms with indices
Simplify non-linear algebraic expressions
Distributive law III
Simplify non-linear algebraic expressions involving distributive law
Expand binomial expressions
Factor algebraic terms
Simplify further algebraic expressions I
Expand further binomial expressions
Expand perfect squares
Expand difference of two squares
Factorise algebraic factors
Simplify algebraic fractions
Add and subtract algebraic fractions
Multiply and divide algebraic fractions
Mixed Operations with Algebraic Fractions
Simplify further algebraic expressions II
Expand further algebraic expressions
What's a binomial?
We've already come across binomial expressions when we looked at how to expand brackets. Expressions such as $2\left(x-3\right)$2(x−3) are the product of a term (outside the brackets) and a binomial expression (the sum or difference of two terms). So a binomial is a mathematical expression in which two terms are added or subtracted. They are usually surrounded by brackets or parentheses, such as ($x+2$x+2).
Recall that to expand $2\left(x-3\right)$2(x−3) we use the distributive property: $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC
Now we want to look at how to multiply two binomials together, such as $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d).
Multiplying Two Binomials
When we multiply binomials of the form $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) we can treat the second binomial $\left(cx+d\right)$(cx+d) as a constant term and apply the distributive property in the form $\left(B+C\right)\left(A\right)=BA+CA$(B+C)(A)=BA+CA. The picture below shows this in action:
As you can see in the picture, we end up with two expressions of the form $A\left(B+C\right)$A(B+C). We can expand these using the distributive property again to arrive at the final answer:
$ax\left(cx+d\right)+b\left(cx+d\right)$ax(cx+d)+b(cx+d) $=$= $acx^2+adx+bcx+bd$acx2+adx+bcx+bd
$=$= $acx^2+\left(ad+bc\right)x+bd$acx2+(ad+bc)x+bd
Let's try an example.
Expand and simplify $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) .
Think: We need to multiply both terms inside $\left(x+5\right)$(x+5) by both terms inside $\left(x+2\right)$(x+2).
$\left(x+5\right)\left(x+2\right)$(x+5)(x+2) $=$= $x\left(x+2\right)+5\left(x+2\right)$x(x+2)+5(x+2)
$=$= $x^2+2x+5x+10$x2+2x+5x+10
$=$= $x^2+7x+10$x2+7x+10
A Visual Interpretation
Let's take the same example as above, $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) and see how this expansion works diagramatically by finding the area of a rectangle.
Notice that the length of the rectangle is $x+5$x+5 and the width is $x+2$x+2. So one expression for the area would be $\left(x+5\right)\left(x+2\right)$(x+5)(x+2).
Another way to express the area would be to split the large rectangle into two smaller rectangles. This way, the area would be $x\left(x+2\right)+5\left(x+2\right)$x(x+2)+5(x+2). Notice that this is the same expression we get after using the distributive property as shown above.
Finally, if we add up the individual parts of this rectangle, we get $x^2+5x+2x+10$x2+5x+2x+10, which simplifies to $x^2+7x+10$x2+7x+10 - the same final answer we found above.
Expand and simplify the following:
$\left(x+2\right)\left(x+5\right)$(x+2)(x+5)
Reveal Solution
$\left(7w+5\right)\left(5w+2\right)$(7w+5)(5w+2)
$-\left(x+5\right)\left(x+2\right)$−(x+5)(x+2)
Calculate $86\times58$86×58 by first expressing it in the form $\left(80+6\right)\left(60-2\right)$(80+6)(60−2).
NA6-5
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
Apply algebraic procedures in solving problems