krit.club logo

Number and Algebra - Algebraic manipulation (expansion, factorization)

Grade 10IB

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

🔑Concepts

Expansion using the Distributive Law: This process involves multiplying a term outside a bracket by every term inside, such as a(b+c)=ab+aca(b + c) = ab + ac. For binomials like (a+b)(c+d)(a + b)(c + d), we use the FOIL method (First, Outer, Inner, Last). Visually, expansion can be viewed as finding the total area of a large rectangle that has been partitioned into smaller sections.

Factorization by Highest Common Factor (HCF): This is the reverse of expansion. It involves identifying the largest factor that divides all terms in an expression and placing it outside a set of brackets. For example, in 4x2+8x4x^2 + 8x, the HCF is 4x4x, resulting in 4x(x+2)4x(x + 2). Visually, this is like identifying a common dimension shared by several rectangular blocks.

Difference of Two Squares (DOTS): An expression in the form a2b2a^2 - b^2 can always be factorized into (ab)(a+b)(a - b)(a + b). Visually, this represents taking a square of area a2a^2, removing a smaller square of area b2b^2 from its corner, and rearranging the remaining L-shaped region into a single rectangle with sides (ab)(a - b) and (a+b)(a + b).

Perfect Square Trinomials: These are specific patterns where (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 and (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2. Visually, (a+b)2(a+b)^2 represents a square with side length a+ba+b, which is composed of a square of area a2a^2, a square of area b2b^2, and two identical rectangles each with area abab.

Factorizing Quadratic Trinomials (x2+bx+cx^2 + bx + c): To factorize a trinomial where the leading coefficient is 11, find two numbers that multiply to give the constant term cc and add to give the coefficient of the middle term bb. These numbers, pp and qq, allow the expression to be written as (x+p)(x+q)(x + p)(x + q). On a coordinate plane, these values relate to the xx-intercepts of the parabola.

Factorization by Grouping: This technique is used for expressions with four terms, such as ax+ay+bx+byax + ay + bx + by. Terms are grouped into pairs, a common factor is extracted from each pair, and then the common binomial factor is extracted. For example, a(x+y)+b(x+y)a(x + y) + b(x + y) becomes (a+b)(x+y)(a + b)(x + y). This can be visualized as splitting a complex composite shape into two simpler rectangular groups.

Simplifying Algebraic Fractions: Algebraic fractions are simplified by first factorizing the numerator and the denominator completely. Once factorized, common binomial or monomial factors that appear in both the top and bottom can be cancelled out (divided to equal 11). This is visually similar to reducing a numerical fraction by dividing out common prime factors.

📐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 the expression: (2x3)(x+5)(x24)(2x - 3)(x + 5) - (x^2 - 4)

Solution:

Step 1: Expand the first two brackets using FOIL: (2x)(x)+(2x)(5)+(3)(x)+(3)(5)=2x2+10x3x15=2x2+7x15(2x)(x) + (2x)(5) + (-3)(x) + (-3)(5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15 Step 2: Distribute the negative sign into the second bracket: (x24)=x2+4-(x^2 - 4) = -x^2 + 4 Step 3: Combine all terms: 2x2+7x15x2+42x^2 + 7x - 15 - x^2 + 4 Step 4: Simplify by combining like terms: $(2x^2 - x^2) + 7x + (-15 + 4) = x^2 + 7x - 11$$

Explanation:

The solution first applies the distributive property to the binomial product and then carefully handles the subtraction of the second expression by reversing the signs of the terms inside the parentheses before combining like terms.

Problem 2:

Factorize the expression completely: 3x2123x^2 - 12

Solution:

Step 1: Identify and factor out the Highest Common Factor (HCF): 3(x24)3(x^2 - 4) Step 2: Recognize that x24x^2 - 4 is a difference of two squares since 4=224 = 2^2: 3(x222)3(x^2 - 2^2) Step 3: Apply the DOTS formula a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b): 3(x2)(x+2)3(x - 2)(x + 2) Final Answer: 3(x2)(x+2)3(x - 2)(x + 2)

Explanation:

The expression is factorized in two stages: first by removing the common numerical factor, and then by applying the Difference of Two Squares identity to the resulting quadratic expression.