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Polynomials - Algebraic Identities

Grade 9CBSE

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

🔑Concepts

Algebraic Identity: An algebraic identity is an equality that holds true for all possible values of its variables. Unlike a standard equation which might only be true for specific values, an identity represents a fundamental property of algebraic operations.

Geometric Visualization of (x+y)2(x + y)^2: This identity can be visualized as the area of a large square with side length (x+y)(x + y). The total area is partitioned into four sections: one square of area x2x^2, another square of area y2y^2, and two identical rectangles each with area xyxy. Summing these parts gives x2+2xy+y2x^2 + 2xy + y^2.

Difference of Two Squares: The expression x2y2x^2 - y^2 represents the area remaining when a square of side yy is removed from a larger square of side xx. This L-shaped area can be sliced and rearranged into a single rectangle with dimensions (x+y)(x + y) and (xy)(x - y), proving the identity x2y2=(x+y)(xy)x^2 - y^2 = (x + y)(x - y).

Product of Binomials with Common Term: The identity (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab is used when two binomials share the same first term. Visually, it represents a rectangle with sides (x+a)(x+a) and (x+b)(x+b), where the area is the sum of a square x2x^2, two rectangles with areas axax and bxbx, and a small rectangle abab.

Expansion of a Trinomial: When squaring a trinomial (x+y+z)2(x + y + z)^2, the result is the sum of the squares of each individual term plus twice the product of every possible pair of terms. This expands to x2+y2+z2+2xy+2yz+2zxx^2 + y^2 + z^2 + 2xy + 2yz + 2zx.

Cubic Identities: Identities like (x+y)3(x + y)^3 and (xy)3(x - y)^3 involve terms of the third degree. Visually, (x+y)3(x + y)^3 represents the volume of a cube with side length (x+y)(x + y), which can be broken down into two smaller cubes and six rectangular prisms (blocks) of varying dimensions.

Conditional Identity: A special case exists for the identity x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx). If the sum of the variables is zero, i.e., x+y+z=0x + y + z = 0, then the entire right side becomes zero, leading to the useful property 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

(xy)2=x22xy+y2(x - y)^2 = x^2 - 2xy + y^2

x2y2=(x+y)(xy)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)

(xy)3=x3y33xy(xy)(x - y)^3 = x^3 - y^3 - 3xy(x - y)

x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

💡Examples

Problem 1:

Evaluate the product 103×107103 \times 107 without multiplying directly using algebraic identities.

Solution:

Step 1: Express the numbers in the form (x+a)(x+b)(x + a)(x + b). We can write 103=100+3103 = 100 + 3 and 107=100+7107 = 100 + 7. \ Step 2: Identify the identity to use: (x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab. \ Step 3: Substitute x=100x = 100, a=3a = 3, and b=7b = 7 into the identity: \ (100+3)(100+7)=(100)2+(3+7)100+(3)(7)(100 + 3)(100 + 7) = (100)^2 + (3 + 7)100 + (3)(7) \ Step 4: Calculate the values: \ =10000+(10)100+21= 10000 + (10)100 + 21 \ =10000+1000+21= 10000 + 1000 + 21 \ =11021= 11021.

Explanation:

We use the identity for the product of two binomials with a common term (x=100x = 100) to simplify the calculation into basic addition and squaring of powers of 10.

Problem 2:

Factorize the expression: 8x3+y3+27z318xyz8x^3 + y^3 + 27z^3 - 18xyz.

Solution:

Step 1: Rewrite the expression in the form a3+b3+c33abca^3 + b^3 + c^3 - 3abc. \ Here, 8x3=(2x)38x^3 = (2x)^3, y3=(y)3y^3 = (y)^3, and 27z3=(3z)327z^3 = (3z)^3. \ The term 18xyz18xyz can be written as 3(2x)(y)(3z)3(2x)(y)(3z). \ Step 2: Use the identity: a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). \ Step 3: Substitute a=2xa = 2x, b=yb = y, and c=3zc = 3z: \ =(2x+y+3z)((2x)2+(y)2+(3z)2(2x)(y)(y)(3z)(3z)(2x))= (2x + y + 3z)((2x)^2 + (y)^2 + (3z)^2 - (2x)(y) - (y)(3z) - (3z)(2x)) \ Step 4: Simplify the second bracket: \ =(2x+y+3z)(4x2+y2+9z22xy3yz6zx)= (2x + y + 3z)(4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6zx).

Explanation:

This problem requires recognizing the structure of the cubic identity involving three variables. By identifying the cube roots of the first three terms, we can apply the factorization formula directly.