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Exploring Algebraic Identities - Factorise algebraic expressions correctly using standard and derived identities

Grade 9CBSE

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

🔑Concepts

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The Factor Theorem states that for a polynomial p(x)p(x) of degree n≥1n \geq 1, if p(a)=0p(a) = 0 for a real number aa, then (x−a)(x - a) is a factor of p(x)p(x). Conversely, if (x−a)(x - a) is a factor, then p(a)=0p(a) = 0. Visually, if you plot the polynomial on a graph, the factor (x−a)(x-a) corresponds to the point where the curve crosses the x-axis at x=ax = a.

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Factorisation by splitting the middle term is used for quadratic polynomials of the form ax2+bx+cax^2 + bx + c. We look for two numbers pp and qq such that p+q=bp + q = b and pq=acpq = ac. This technique is like breaking a large rectangular area representing the polynomial into smaller, manageable sections that share common side lengths.

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Factorisation by grouping involves rearranging terms of a polynomial into groups that have a common factor. For example, in ax+ay+bx+byax + ay + bx + by, we group as (ax+ay)+(bx+by)(ax + ay) + (bx + by), factor out common terms to get a(x+y)+b(x+y)a(x + y) + b(x + y), and finally (a+b)(x+y)(a + b)(x + y). Imagine organizing a set of tiles into rows and columns where the 'common factor' represents a shared dimension.

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The difference of two squares x2−y2x^2 - y^2 is factorised into (x−y)(x+y)(x - y)(x + y). This can be visualized as taking a square with side xx, removing a smaller square with side yy from the corner, and then slicing and rearranging the remaining shape to form a rectangle with dimensions (x+y)(x + y) and (x−y)(x - y).

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For cubic polynomials like ax3+bx2+cx+dax^3 + bx^2 + cx + d, factorisation often begins with the Factor Theorem to find one linear factor (x−k)(x - k) by testing divisors of the constant term dd. Once one factor is found, the polynomial is divided by it to obtain a quadratic polynomial, which is then factorised further.

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Perfect square trinomials x2+2xy+y2x^2 + 2xy + y^2 and x2−2xy+y2x^2 - 2xy + y^2 are factorised into (x+y)2(x + y)^2 and (x−y)2(x - y)^2 respectively. Geometrically, x2+2xy+y2x^2 + 2xy + y^2 represents the total area of a large square made of two smaller squares (x2x^2 and y2y^2) and two identical rectangles (each with area xyxy).

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Algebraic identities involving three variables, such as x3+y3+z3−3xyzx^3 + y^3 + z^3 - 3xyz, can be factorised into (x+y+z)(x2+y2+z2−xy−yz−zx)(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx). This identity is particularly useful when x+y+z=0x + y + z = 0, leading to the result x3+y3+z3=3xyzx^3 + y^3 + z^3 = 3xyz.

📐Formulae

(x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2

(x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2

x2−y2=(x+y)(x−y)x^2 - y^2 = (x + y)(x - y)

(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

(x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

(x+y)3=x3+y3+3xy(x+y)(x + y)^3 = x^3 + y^3 + 3xy(x + y)

(x−y)3=x3−y3−3xy(x−y)(x - y)^3 = x^3 - y^3 - 3xy(x - y)

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

💡Examples

Problem 1:

Factorise the quadratic polynomial 12x2−7x+112x^2 - 7x + 1 by splitting the middle term.

Solution:

Step 1: Identify a=12a = 12, b=−7b = -7, and c=1c = 1. We need two numbers pp and qq such that p+q=−7p + q = -7 and pq=12×1=12pq = 12 \times 1 = 12.\nStep 2: The numbers are −3-3 and −4-4, since (−3)+(−4)=−7(-3) + (-4) = -7 and (−3)×(−4)=12(-3) \times (-4) = 12.\nStep 3: Split the middle term: 12x2−3x−4x+112x^2 - 3x - 4x + 1.\nStep 4: Group the terms: (12x2−3x)−(4x−1)(12x^2 - 3x) - (4x - 1).\nStep 5: Factor out common terms: 3x(4x−1)−1(4x−1)3x(4x - 1) - 1(4x - 1).\nStep 6: Final factorisation: (3x−1)(4x−1)(3x - 1)(4x - 1).

Explanation:

This approach reduces a three-term quadratic into a four-term expression that can be factorised by grouping common terms from the first two and last two segments.

Problem 2:

Factorise x3−23x2+142x−120x^3 - 23x^2 + 142x - 120 using the Factor Theorem.

Solution:

Step 1: Let p(x)=x3−23x2+142x−120p(x) = x^3 - 23x^2 + 142x - 120. Look for factors of 120120 (e.g., ±1,±2\pm 1, \pm 2).\nStep 2: Test x=1x = 1: p(1)=13−23(1)2+142(1)−120=1−23+142−120=0p(1) = 1^3 - 23(1)^2 + 142(1) - 120 = 1 - 23 + 142 - 120 = 0. Since p(1)=0p(1) = 0, (x−1)(x - 1) is a factor.\nStep 3: Divide p(x)p(x) by (x−1)(x - 1) using long division or synthetic division to get the quotient: x2−22x+120x^2 - 22x + 120.\nStep 4: Factorise the quadratic x2−22x+120x^2 - 22x + 120 by splitting the middle term: x2−12x−10x+120=x(x−12)−10(x−12)=(x−10)(x−12)x^2 - 12x - 10x + 120 = x(x - 12) - 10(x - 12) = (x - 10)(x - 12).\nStep 5: Combine all factors: (x−1)(x−10)(x−12)(x - 1)(x - 10)(x - 12).

Explanation:

For higher-degree polynomials, we find one root via trial and error based on the Factor Theorem and then simplify the remaining polynomial using division.