# Algebraic Expressions and Identities - Class 8 - Mathematics

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## Examples - Algebraic Expressions and Identities | NCERT | Mathematics | Class 8

Add: $7 x y+5 y z-3 z x, 4 y z+9 z x-4 y,-3 x z+5 x-2 x y$

To add the given polynomials, $7xy + 5yz - 3zx, 4yz + 9zx - 4y, -3xz + 5x - 2xy$, follow these steps:

**Identify like terms:**Like terms are those that contain the same variables raised to the same power. In this context, terms involving $xy$, $yz$, $zx$ (or equivalently $xz$), and single variable terms $x$ and $y$ can be combined.**Combine like terms:**For $xy$: Combine $7xy$ and $-2xy$ to get $5xy$.

For $yz$: Combine $5yz$ and $4yz$ to get $9yz$.

For $zx$ (same as $xz$): Combine $-3zx$, $9zx$, and $-3xz$ to get $3zx$.

For $x$: You only have $5x$, so it remains as it is.

For $y$: You only have $-4y$, so it remains as it is.

**Sum up the results:**Finally, sum up the combined like terms to get the total sum: $$5xy + 9yz + 3zx + 5x - 4y$$

So, the sum of $7xy + 5yz - 3zx$, $4yz + 9zx - 4y$, and $-3xz + 5x - 2xy$ is $5xy + 9yz + 3zx + 5x - 4y$.

Subtract $5 x^{2}-4 y^{2}+6 y-3$ from $7 x^{2}-4 x y+8 y^{2}+5 x-3 y$.

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Sign up now## Exercise 8.1 - Algebraic Expressions and Identities | NCERT | Mathematics | Class 8

Add the following

(i) $a b-b c, b c-c a, c a-a b$

(ii) $a-b+a b, b-c+b c, c-a+a c$

(iii) $2 p^{2} q^{2}-3 p q+4,5+7 p q-3 p^{2} q^{2}$

(iv) $l^{2}+m^{2}, m^{2}+n^{2}, n^{2}+l^{2}$

$2 l m+2 m n+2 n l$

The results of the column-wise additions are:

(i) $0$

(ii) $ab + ac + b*c$

(iii) $9 + 4pq - p^2*q^2$

(iv) $2l^2 + 2lm + 2m^2 + 2ln + 2mn + 2*n^2$

Subtract $4 a-7 a b+3 b+12$ from $12 a-9 a b+5 b-3$

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Sign up nowSubtract $3 x y+5 y z-7 z x$ from $5 x y-2 y z-2 z x+10 x y z$

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Sign up nowSubtract $4 p^{2} q-3 p q+5 p q^{2}-8 p+7 q-10$ from $18-3 p-11 q+5 p q-2 p q^{2}+5 p^{2} q$

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## Exercise 8.2 - Algebraic Expressions and Identities | NCERT | Mathematics | Class 8

Find the product of the following pairs of monomials.

(i) $4,7 p$

(ii) $-4 p, 7 p$

(iii) $-4 p, 7 p q$

(iv) $4 p^{3},-3 p$

(v) $4 p, 0$

The products of the given pairs of monomials are:

(i) $28p$

(ii) $-28p^2$

(iii) $-28p^2q$ (Simplifying $-28 * p * pq$ gives $-28p^2q$)

(iv) $-12p^4$

(v) $0$

Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$(p, q) ;(10 m, 5 n) ;\left(20 x^{2}, 5 y^{2}\right) ;\left(4 x, 3 x^{2}\right) ;(3 m n, 4 n p)$

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Complete the table of products.

$\frac{\text{First monomial} \rightarrow}{\text{Second monomial} \downarrow}$ | $2x$ | $-5y$ | $3x^2$ | $-4xy$ | $7x^2y$ | $-9x^2y^2$ |
---|---|---|---|---|---|---|

$2x$ | $4x^2$ | |||||

$-5y$ | ||||||

2$ | $-15x^2y$ | |||||

4xy$ | ||||||

$7x^2y$ | ||||||

$-9 x^{2} y^{2}$ |

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Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) $5 a, 3 a^{2}, 7 a^{4}$

(ii) $2 p, 4 q, 8 r$

(iii) $x y, 2 x^{2} y, 2 x y^{2}$

(iv) $a, 2 b, 3 c$.

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Obtain the product of

(i) $x y, y z, z x$

(ii) $a,-a^{2}, a^{3}$

(iii) $2,4 y, 8 y^{2}, 16 y^{3}$

(iv) $a, 2 b, 3 c, 6 a b c$

(v) $m,-m n, m n p$

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## Exercise 8.3 - Algebraic Expressions and Identities | NCERT | Mathematics | Class 8

Carry out the multiplication of the expressions in each of the following pairs.

(i) $4 p, q+r$

(ii) $a b, a-b$

(iii) $a+b, 7 a^{2} b^{2}$

(iv) $a^{2}-9,4 a$

(v) $p q+q r+r p, 0$

Here are the results of carrying out the multiplication of the expressions in each of the given pairs:

$(i) , 4p \times (q+r) = 4pq + 4pr$

$(ii) , ab \times (a-b) = a^2b - ab^2$

$(iii) , (a+b) \times 7a^2b^2 = 7a^3b^2 + 7a^2b^3$

$(iv) , (a^2-9) \times 4a = -36a + 4a^3$

$(v) , (pq+qr+rp) \times 0 = 0$

Complete the table.

First expression | Second expression | Product | |
---|---|---|---|

(i) | $a$ | $b+c+d$ | |

(ii) | $x+y-5$ | $5 x y$ | |

(iii) | $p$ | $6 p^{2}-7 p+5$ | |

(iv) | $4 p^{2} q^{2}$ | $p^{2}-q^{2}$ | |

(v) | $a+b+c$ | $a b c$ |

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Find the product.

(i) $\left(a^{2}\right) \times\left(2 a^{22}\right) \times\left(4 a^{26}\right)$

(ii) $\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)$

(iii) $\left(-\frac{10}{3} p q^{3}\right) \times\left(\frac{6}{5} p^{3} q\right)$

(iv) $x \times x^{2} \times x^{3} \times x^{4}$

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(a) Simplify $3 x(4 x-5)+3$ and find its values for

(i) $x=3$

(ii) $x=\frac{1}{2}$.

(b) Simplify $a\left(a^{2}+a+1\right)+5$ and find its value for

(i) $a=0$

(ii) $a=1$

(iii) $a=-1$.

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(a) Add: $p(p-q), q(q-r)$ and $r(r-p)$

(b) Add: $2 x(z-x-y)$ and $2 y(z-y-x)$

(c) Subtract: $3 l(l-4 m+5 n)$ from $4 l(10 n-3 m+2 l)$

(d) Subtract: $3 a(a+b+c)-2 b(a-b+c)$ from $4 c(-a+b+c)$

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## Extra Questions - Algebraic Expressions and Identities | NCERT | Mathematics | Class 8

Evaluate:

$$ (4pq + 3q)^2 - (4pq - 3q)^2 $$

(A) $48pq^2$

(B) $48p^2q$

(C) $48p^2q^2$

(D) $48pq$

The correct option is **(A) $48pq^2$**

We start by recognizing the given expression as a difference of squares. The expression can be written as: $$ (4pq + 3q)^2 - (4pq - 3q)^2 $$ This can be simplified using the identity for the difference of squares, which is: $$ (a^2 - b^2) = (a+b)(a-b) $$ Here, let $a = 4pq + 3q$ and $b = 4pq - 3q$.

Applying the identity: $$ [(4pq + 3q) - (4pq - 3q)] [(4pq + 3q) + (4pq - 3q)] $$ Simplifying inside the brackets: $$ [4pq + 3q - 4pq + 3q][4pq + 3q + 4pq - 3q] $$ This reduces to: $$ [6q][8pq] $$ Thus, multiplying these, we get: $$ 48pq^2 $$

**Therefore, the correct answer is indeed $(A)$ $48pq^2$.**

Divide $x^{4a} + x^{2a}y^{2b} + y^{4b}$ by $x^{2a} + x^{a}y^{b} + y^{2b}$