NCERT Class 9 Maths: Polynomials Exercise 2.5 – Algebraic Identities Simplified

Understanding Algebraic Identities for Polynomials

Algebraic identities are equations that are true for all values of the variables involved. In Chapter 2, Exercise 2.5, we encounter several fundamental identities that are indispensable for manipulating polynomials. These identities provide a systematic way to expand products of binomials and trinomials, and conversely, to factorize certain types of polynomial expressions. Instead of tediously multiplying term by term, which can be prone to errors and time-consuming, identities offer elegant solutions. For instance, expanding $(x+y)^2$ directly means $(x+y) \times (x+y) = x(x+y) + y(x+y) = x^2+xy+yx+y^2 = x^2+2xy+y^2$. The identity $(x+y)^2 = x^2 + 2xy + y^2$ gives us this result instantly. Similarly, factorization, the reverse process, becomes much simpler. Recognizing patterns like $x^2 - y^2$ immediately tells us it can be factorized as $(x-y)(x+y)$. Mastery of these identities is not just about memorization, but understanding when and how to apply them effectively to solve problems related to expansion, factorization, and evaluation. This section of polynomial ex 2 5 class 9 ncert builds a strong foundation for your future algebraic studies.

Essential Identities Covered in Ex 2.5

Identity I: $(x+y)^2$

$(x+y)^2 = x^2 + 2xy + y^2$. This identity is used to expand the square of a sum of two terms.

Identity II: $(x-y)^2$

$(x-y)^2 = x^2 - 2xy + y^2$. This identity is used to expand the square of a difference of two terms.

Identity III: $x^2 - y^2$

$x^2 - y^2 = (x+y)(x-y)$. This identity is for factorizing the difference of two squares.

Identity IV: $(x+a)(x+b)$

$(x+a)(x+b) = x^2 + (a+b)x + ab$. This identity helps expand products of two linear binomials with a common term 'x'.

Identity V: $(x+y+z)^2$

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$. This identity expands the square of a trinomial.

Identity VI: $(x+y)^3$

$(x+y)^3 = x^3 + y^3 + 3xy(x+y) = x^3 + 3x^2y + 3xy^2 + y^3$. This expands the cube of a sum of two terms.

Identity VII: $(x-y)^3$

$(x-y)^3 = x^3 - y^3 - 3xy(x-y) = x^3 - 3x^2y + 3xy^2 - y^3$. This expands the cube of a difference of two terms.

Identity VIII: $x^3 + y^3 + z^3 - 3xyz$

$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. A powerful identity for factorizing and simplifying cubic expressions. A special case applies if $x+y+z=0$, then $x^3+y^3+z^3 = 3xyz$.

Step-by-Step Application of Identities

• Example 1: Expanding using (x+a)(x+b) Problem: Expand $(x+4)(x+10)$. Solution: Step 1: Identify the identity to be used. Here, it is Identity IV: $(x+a)(x+b) = x^2 + (a+b)x + ab$. Step 2: Compare the given expression with the identity. We have $a=4$ and $b=10$. Step 3: Substitute the values into the identity. $(x+4)(x+10) = x^2 + (4+10)x + (4)(10)$ Step 4: Simplify the expression. $= x^2 + 14x + 40$ Final answer: $x^2 + 14x + 40$
• Example 2: Factorizing using (x+y+z)^2 Problem: Factorize $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$. Solution: Step 1: Recognize the pattern as Identity V: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$. Note: We use 'a, b, c' here to avoid confusion with 'x, y, z' in the expression. Step 2: Rewrite the terms as squares. $4x^2 = (2x)^2$, $9y^2 = (3y)^2$, $16z^2 = (-4z)^2$ (we need to consider signs for the cross products). Step 3: Check the cross terms: $2ab = 2(2x)(3y) = 12xy$ (Matches) $2bc = 2(3y)(-4z) = -24yz$ (Matches) $2ca = 2(-4z)(2x) = -16xz$ (Matches) Step 4: Apply the identity by writing the sum of the terms being squared. The expression is of the form $(2x + 3y - 4z)^2$. Final answer: $(2x + 3y - 4z)^2$
• Example 3: Evaluating products without direct multiplication Problem: Evaluate $103 \times 107$ without multiplying directly. Solution: Step 1: Express the numbers in a form suitable for an identity. We can write $103 = 100+3$ and $107 = 100+7$. Step 2: Identify the identity. This fits Identity IV: $(x+a)(x+b) = x^2 + (a+b)x + ab$. Step 3: Assign values. Here, $x=100$, $a=3$, and $b=7$. Step 4: Substitute into the identity. $103 \times 107 = (100+3)(100+7) = 100^2 + (3+7)(100) + (3)(7)$ Step 5: Perform the calculations. $= 10000 + (10)(100) + 21$ $= 10000 + 1000 + 21$ $= 11021$ Final answer: $11021$

Common Mistakes and YoLearn Exam Tips

Navigating polynomial ex 2 5 class 9 ncert problems requires attention to detail. Here are some common pitfalls and tips to avoid them:

1. Sign Errors: This is perhaps the most frequent mistake. Always be careful with negative signs, especially when applying identities like $(x-y)^2$ or identities involving negative terms within parentheses. For example, $(x-y)^2 = x^2 - 2xy + y^2$, not $x^2 + 2xy - y^2$. Similarly, when using $(x+y+z)^2$, if $y$ or $z$ is negative, ensure the $2xy$, $2yz$, $2zx$ terms reflect that sign. For instance, $(x-y+z)^2 = x^2 + (-y)^2 + z^2 + 2x(-y) + 2(-y)z + 2zx = x^2+y^2+z^2-2xy-2yz+2zx$.

1. Incorrect Application of Identities: Students sometimes confuse identities or try to force an expression into an incorrect identity. For example, $(x+y)^3$ is not simply $x^3+y^3$. Remember the middle terms: $3x^2y$ and $3xy^2$. Always ensure the given expression perfectly matches the structure of the chosen identity before applying it.

1. Forgetting the Special Case for Identity VIII: Remember that if $x+y+z = 0$, then Identity VIII simplifies to $x^3+y^3+z^3 = 3xyz$. This is a very common type of question in exams and can save a lot of calculation if you spot it.

1. Incomplete Factorization: When asked to factorize, ensure you factorize completely. Sometimes, after applying one identity, the resulting expression might still be factorizable using another identity or by taking out common factors.

YoLearn Exam Tip: Always write down the identity you are using before applying it. This helps clarify your thought process and minimizes errors. Practice identifying the correct identity quickly by looking for specific patterns (e.g., sum of squares, difference of squares, cube of a sum/difference).

Practice Questions with Solutions

• Q: Expand $(3a+4b+5c)^2$. A: Step 1: Identify the identity. This is of the form $(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$. Step 2: Assign $x=3a$, $y=4b$, $z=5c$. Step 3: Substitute the values into the identity. $(3a+4b+5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)$ Step 4: Simplify each term. $= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca Final answer: $9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca$
• Q: Factorize $27y^3 + 125z^3$. A: Step 1: Recognize this as a sum of two cubes, which fits the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$. Step 2: Express the terms as cubes. $27y^3 = (3y)^3$ and $125z^3 = (5z)^3$. Step 3: Assign $a=3y$ and $b=5z$. Step 4: Substitute into the identity. $(3y)^3 + (5z)^3 = (3y+5z)((3y)^2 - (3y)(5z) + (5z)^2)$ Step 5: Simplify the terms inside the second bracket. $= (3y+5z)(9y^2 - 15yz + 25z^2) Final answer: $(3y+5z)(9y^2 - 15yz + 25z^2)$
• Q: Evaluate $99^3$ using a suitable identity. A: Step 1: Express $99$ in a form suitable for an identity, such as $100-1$. Step 2: Identify the identity. This is of the form $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$. Step 3: Assign $x=100$ and $y=1$. Step 4: Substitute into the identity. $(100-1)^3 = 100^3 - 1^3 - 3(100)(1)(100-1)$ Step 5: Perform the calculations. $= 1000000 - 1 - 300(99)$ $= 1000000 - 1 - 29700$ $= 970299 Final answer: $970299$
• Q: If $x+y+z=0$, show that $x^3+y^3+z^3 = 3xyz$. A: Step 1: Recall the identity $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Step 2: The problem states that $x+y+z=0$. Step 3: Substitute $x+y+z=0$ into the identity. $x^3+y^3+z^3-3xyz = (0)(x^2+y^2+z^2-xy-yz-zx)$ Step 4: Simplify the right-hand side. Any number multiplied by zero is zero. $x^3+y^3+z^3-3xyz = 0$ Step 5: Transpose $-3xyz$ to the right-hand side. $x^3+y^3+z^3 = 3xyz$ Final answer: Proven.
• Q: Expand $(2x-y+z)^2$. A: Step 1: Use the identity $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$. Step 2: Assign $a=2x$, $b=-y$, $c=z$. Step 3: Substitute into the identity. $(2x-y+z)^2 = (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$ Step 4: Simplify each term carefully, paying attention to signs. $= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx Final answer: $4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$

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