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Algebra - Expanding and factorizing algebraic expressions, including quadratic expressions

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Distributive Law: This concept involves multiplying a single term outside a bracket by every term inside the bracket, represented as a(b+c)=ab+aca(b + c) = ab + ac. Visually, this can be understood as finding the area of a large rectangle with width aa and length (b+c)(b+c), which is equivalent to the sum of the areas of two smaller rectangles, abab and acac.

Expanding Binomials (FOIL): To expand an expression like (a+b)(c+d)(a + b)(c + d), we multiply the First terms, Outer terms, Inner terms, and Last terms. This is often visualized using a 2x2 area model or grid, where each of the four cells contains the product of one term from each binomial, which are then summed together to get ac+ad+bc+bdac + ad + bc + bd.

Common Factorizing: This is the inverse process of expansion. It involves identifying the Highest Common Factor (HCF) of all terms in an expression and placing it outside a set of brackets. For example, in 4x2+8x4x^2 + 8x, the HCF is 4x4x, allowing the expression to be rewritten as 4x(x+2)4x(x + 2). This is like dividing the total area of a rectangle by a known side length to find the remaining side.

Difference of Two Squares: A specific pattern where two perfect squares are subtracted, written as a2b2a^2 - b^2. Visually, if you take a square of area a2a^2 and remove a smaller square of area b2b^2 from its corner, the remaining shape can be sliced and rearranged into a rectangle with side lengths (ab)(a - b) and (a+b)(a + b), giving the identity (ab)(a+b)(a - b)(a + b).

Perfect Square Trinomials: Expanding (a+b)2(a + b)^2 or (ab)2(a - b)^2 results in a2+2ab+b2a^2 + 2ab + b^2 or a22ab+b2a^2 - 2ab + b^2 respectively. Visually, (a+b)2(a+b)^2 represents a square with side length (a+b)(a+b). Inside this square, you will see one square of area a2a^2, one square of area b2b^2, and two rectangles each with area abab.

Factorizing Trinomials (x2+bx+cx^2 + bx + c): To factorize a quadratic where the x2x^2 coefficient is 1, we look for two numbers that multiply to give the constant cc and add to give the coefficient bb. Visually, this is the process of arranging x2x^2 tiles, xx tiles, and unit tiles into a single large rectangle where the side lengths represent the factors (x+p)(x+p) and (x+q)(x+q).

Factorizing by Grouping: This technique is applied to expressions with four terms, such as ax+ay+bx+byax + ay + bx + by. We group the terms into two pairs that share common factors: a(x+y)+b(x+y)a(x + y) + b(x + y). Since (x+y)(x + y) is now a common factor for both groups, it can be factored out to produce (a+b)(x+y)(a + b)(x + y).

📐Formulae

a(b+c)=ab+aca(b + c) = ab + ac

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

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2

a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b)

x2+(p+q)x+pq=(x+p)(x+q)x^2 + (p+q)x + pq = (x + p)(x + q)

💡Examples

Problem 1:

Expand and simplify: (2x3)(x+4)(2x - 3)(x + 4)

Solution:

Step 1: Multiply the First terms: 2xx=2x22x \cdot x = 2x^2 Step 2: Multiply the Outer terms: 2x4=8x2x \cdot 4 = 8x Step 3: Multiply the Inner terms: 3x=3x-3 \cdot x = -3x Step 4: Multiply the Last terms: 34=12-3 \cdot 4 = -12 Step 5: Combine all terms: 2x2+8x3x122x^2 + 8x - 3x - 12 Step 6: Simplify like terms: 2x2+5x122x^2 + 5x - 12

Explanation:

We use the FOIL method to distribute each term in the first binomial across the second. Finally, we combine the middle linear terms (8x8x and 3x-3x) to reach the simplest form.

Problem 2:

Factorize the quadratic expression: x25x14x^2 - 5x - 14

Solution:

Step 1: Identify the constant c=14c = -14 and the middle coefficient b=5b = -5. Step 2: Find two numbers that multiply to 14-14 and add to 5-5. The pairs for 14-14 are (1,14),(1,14),(2,7),(2,7)(1, -14), (-1, 14), (2, -7), (-2, 7). Step 3: Check sums: 2+(7)=52 + (-7) = -5. The numbers are 22 and 7-7. Step 4: Write in factorized form: (x+2)(x7)(x + 2)(x - 7)

Explanation:

To factorize this trinomial, we look for two integers whose product is the constant term and whose sum is the coefficient of xx. We then place these integers into the binomial factor template (x+p)(x+q)(x + p)(x + q).