Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Expansion is the algebraic process of removing brackets by multiplying terms. Visually, the expansion can be represented as the area of a large rectangle with sides and , which is equal to the sum of the areas of four smaller internal rectangles: and .
Squaring a binomial results in a trinomial. Geometrically, this represents a square of side being divided into four sections: a square of area , a square of area , and two identical rectangles each of area . The result is .
The product of a sum and a difference, , leads to the 'Difference of Two Squares' identity. Visually, if you take a square with side (area ) and remove a smaller square of side (area ) from the corner, the remaining shape can be sliced and rearranged into a rectangle with dimensions and .
Expansion of a trinomial involves the sum of the squares of each term plus twice the product of every possible pair of terms. This expands into 6 terms: .
Cubing a binomial represents the volume of a cube with side length . This volume is composed of a smaller cube , another cube , three rectangular prisms with dimensions , and three rectangular prisms with dimensions .
Expansion of reciprocals involves expressions like . Because the product of a number and its reciprocal is 1 (), the middle term of the expansion becomes a constant integer, which is a unique property used to solve higher-order power problems.
Conditional identities are specific results that hold true under given constraints. A common one is: if , then . This simplifies calculations for sum of cubes significantly when the sum of the bases is zero.
📐Formulae
or
or
💡Examples
Problem 1:
Expand and find the value if .
Solution:
- Use the identity . \ 2. Let and . \ 3. \ 4. \ 5. Substitute : \ 6. .
Explanation:
We apply the square of a binomial formula. Note that both the coefficient and the variable must be squared inside the parentheses.
Problem 2:
If , find the value of and .
Solution:
- Square the given equation: \ 2. \ 3. \ 4. Square the result: \ 5. \ 6. .
Explanation:
By squaring the sum of a term and its reciprocal, the middle term simplifies to the constant 2, allowing us to find the sum of squares by simple subtraction.