CBSE Class 8 Maths Exemplar Chapter 7: Algebraic Expressions, Identities and Factorization NCERT Solutions

Let's consider a monomial, for example, $2x$, and a binomial, for example, $(x + y)$. To find their product, we use the distributive property:

$$2x\; \backslash times\; (x\; +\; y)\; =\; 2x\; \backslash times\; x\; +\; 2x\; \backslash times\; y$$

This simplifies to:

$$=\; 2x^2\; +\; 2xy$$

The resulting expression, $2x^2 + 2xy$, consists of two terms. Therefore, the product of a monomial and a binomial is always a binomial.

A polynomial is an algebraic expression consisting of variables and coefficients, where the variables involve only the operations of addition, subtraction, multiplication, and non-negative integer exponents. Therefore, in a polynomial, the exponents of the variables must always be non-negative integers (i.e., 0, 1, 2, 3, ...).

Let's expand the expression $(a - b)^2$:

$(a\; -\; b)^2\; =\; (a\; -\; b)(a\; -\; b)$

Using the distributive property (or FOIL method):

$=\; a(a\; -\; b)\; -\; b(a\; -\; b)$

$=\; a\; \backslash times\; a\; -\; a\; \backslash times\; b\; -\; b\; \backslash times\; a\; +\; b\; \backslash times\; b$

Since multiplication is commutative ($a \times b = b \times a$), we have:

$$=\; a^2\; -\; ab\; -\; ab\; +\; b^2$$

Combining the like terms:

$=\; a^2\; -\; 2ab\; +\; b^2$

Now let's consider $(a + b)^2$:

$$(a\; +\; b)^2\; =\; (a\; +\; b)(a\; +\; b)$$

$=\; a(a\; +\; b)\; +\; b(a\; +\; b)$

$=\; a\; \backslash times\; a\; +\; a\; \backslash times\; b\; +\; b\; \backslash times\; a\; +\; b\; \backslash times\; b$

$=\; a^2\; +\; ab\; +\; ab\; +\; b^2$

$=\; a^2\; +\; 2ab\; +\; b^2$

Comparing these expansions with the given options, we find that option (b) is the correct identity.

To find the sum of $-7pq$ and $2pq$, we add the coefficients of the like terms:

$-7pq\; +\; 2pq$

We can factor out the common term $pq$:

$$=\; pq(-7\; +\; 2)$$

Now, perform the addition of the coefficients:

$=\; -5pq$

Thus, the sum of $-7pq$ and $2pq$ is $-5pq$.

To subtract $-3x^2y^2$ from $x^2y^2$, we set up the subtraction as follows:

$x^2y^2\; -\; (-3x^2y^2)$

Subtracting a negative number is the same as adding its positive counterpart:

$=\; x^2y^2\; +\; 3x^2y^2$

Now, we combine the like terms by adding their coefficients:

$$=\; (1\; +\; 3)\; x^2\; y^2$$

$$=\; 4x^2\; y^2$$

Therefore, subtracting $-3x^2y^2$ from $x^2y^2$ results in $4x^2y^2$.

Like terms are terms that have the same variables raised to the same powers. In the given term $4m^3n^2$, the variable part is $m^3n^2$. We need to find an option that also has the exact same variable part $m^3n^2$.

Let's examine the options:

• (a) $4m^2n^2$ - Variable part is $m^2n^2$ (different powers).
• (b) $-6m^3n^2$ - Variable part is $m^3n^2$ (same powers).
• (c) $6pm^3n^2$ - Variable part is $pm^3n^2$ (includes an extra variable 'p').
• (d) $4m^3n$ - Variable part is $m^3n$ (different power for 'n').

The term $-6m^3n^2$ has the same variable part $m^3n^2$ as the given term $4m^3n^2$. Therefore, $-6m^3n^2$ is a like term.

A binomial is an algebraic expression that contains exactly two unlike terms. Let's analyze each option:

• (a) $7\; \backslash times\; a\; +\; a$: This expression simplifies to $7a + a = 8a$. This is a monomial (one term).
• (b) $6a^2\; +\; 7b\; +\; 2c$: This expression has three unlike terms ($6a^2$, $7b$, and $2c$). This is a trinomial.
• (c) $4a\; \backslash times\; 3b\; \backslash times\; 2c$: This expression simplifies to $24abc$. This is a monomial (one term).
• (d) $6(a^2\; +\; b)$: Using the distributive property, this expression expands to $6 \times a^2 + 6 \times b = 6a^2 + 6b$. This expression has two unlike terms ($6a^2$ and $6b$).

Therefore, the expression $6(a^2\; +\; b)$ is a binomial.