Expansion takes an expression from factored form and uses the distributive property to remove the brackets and write the expression as a sum of products. Such as $\left(a+2\right)\left(b+3\right)=ab+3a+2b+6$(a+2)(b+3)=ab+3a+2b+6.
The distributive property: $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC
This can be used over more terms:
$A\left(B+C+D\right)=AB+AC+AD$A(B+C+D)=AB+AC+AD
Or applied multiple times:
| $\left(A+B\right)\left(C+D\right)$(A+B)(C+D) | $=$= | $A\left(C+D\right)+B\left(C+D\right)$A(C+D)+B(C+D) |
| $=$= | $AC+AD+BC+BD$AC+AD+BC+BD |
You can reduce this working to one line by remembering the first term multiplies each in the second bracket, then the second term multiplies each in the second bracket.
Hence, $\left(x+5\right)\left(x+2\right)=x^2+2x+5x+10$(x+5)(x+2)=x2+2x+5x+10. Which we could then simplify to $x^2+7x+10$x2+7x+10.
After expanding check if the expression can be simplified by collecting like-terms.
Practice questions
Question 1
Expand and simplify the following:
$\left(2n+5\right)\left(5n+2\right)-4$(2n+5)(5n+2)−4
Question 2
Expand and simplify the following expression $\left(x+7\right)\left(x-7\right)-\left(x-3\right)^2$(x+7)(x−7)−(x−3)2
Further expansion
When multiplying more than two expressions reduce the problem to steps and only ever multiply two brackets at a time.
Practice questions
Question 3
Expand the following:
$\left(2x+1\right)\left(5x+7\right)\left(2x-1\right)$(2x+1)(5x+7)(2x−1)
Question 4
Expand $\left(3c+2\right)\left(2c^2+2\right)^2$(3c+2)(2c2+2)2.
Question 5
Consider the expansion of $\left(4x^3-4x^2+ax+b\right)\left(4x^2-2x+9\right)$(4x3−4x2+ax+b)(4x2−2x+9).
In the expansion, the coefficient of $x^3$x3 is $12$12. Form an equation and solve for the value of $a$a.
In the expansion, the coefficient of $x^2$x2 is $4$4. Form an equation and solve for the value of $b$b.