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Worksheet for Algebraic Expressions and Identities Class 8 Maths Chapter 9 CBSE

In this page we have algebraic expressions and identities class 8 worksheets with solutions . Hope you like them and do not forget to like , social share and comment at the end of the page.

Short answer type

Question 1
Use a suitable identity to get each of the following products.
(a) (p - 11) (p + 11)
(b) (2y + 5) (2y - 5)
(c) (12a - 9) (12a +9)
(d) (2a-1/2)(2a-1/2)
(e) (1.1m - 0.4) (1.1m + 0.4)
(f) (a²+ b²) (- a²+ b²)
(g) (6x - 7) (6x + 7)
(h) (- a/2 + c/2) (- a/2 + c/2)
(i) [(p/8)+(3q/4)] [(p/8)+(3q/4)]
(j) (3a + 9b) (3a - 9b)
(k) 2(a - 9)²
(l) 5(xy - 3z)²
(m) (6x+ 5y)²
(n) 36[(3p/2}) + (2q/3)]²
(o) (x - 0.5y)²
(p) (2xy - 5y)²

Answer

We will be using below identities in these question
(a + b)² = a² + 2ab + b²
(a � b)² = a² � 2ab + b²
(a � b)(a + b) = a² � b²

(a) p² -121
(b) 4y² -25
(c) 144a²-81
(d) 4a²+1/4 -2a
(e) 1.21m² -.16
(f) b&sup4; -a&sup4;
(g) 36x² -49
(h) c²/4 - a²/4
(i) p²/64 + 9q²/16 +3pq/16 = (p²+36q²+12pq)/64
(j) 9a²-81b²
(k) 4(a² + 81-18a)
(l) 25(x²y²+9z²-6xyz)
(m) 36x²+25y²+60xy
(n) 36[9p²/4 + 4q²/9 + 2pq]= 9p²+16q² + 64pq
(o) x²+.25y²-xy
(p) 4x²y²+25y²-20xy²

Question 2
Use the identity (x + a) (x + b) = x²+ (a + b) x + ab to find the following products.
(i) (p + 10) (p + 11)
(ii) (4x + 9) (4x + 12)
(iii) (x - 5) (x - 1)
(iv) (9x - 5) (9x - 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a²+ 9) (2a²+ 5)

Answer

(i) p²+21p+110
(ii) 16x² +84x+108
(iii) x² -6x+5
(iv) 81x²-54x+5
(v) 4x² +16xy+15y²
(vi) 4a&sup4;+28a² + 45

Question 3
Simplify the following
(i) (x²- y²)² + 4x²y²
(ii) (p + q)²- (p - q)² + p²q²
(iii) (2m - 8n)²+ (2m + 8n)²
(iv) (4m + 5n)²+ (5m + 4n)² + (4m + 5n) (4m -5n)
(v) (.5p - 1.5q)²- (.5p - 1.5q)² +p²q²
(vi) (ab - bc)²+2ab²c
(vii) (m²- n²m)²+ 2m²n²

Answer

(i) (x²- y²)² + 4x²y²= x⁴+ y⁴-2x²y²+4x²y² =x⁴+ y⁴+2x²y²=(x²+ y²)²
(ii) (p + q)²- (p - q)² + p²q²=4pq+p²q²=pq(4+pq)
(iii) (2m - 8n)²+ (2m + 8n)²=8m²+128n²
(iv) (4m + 5n)²+ (5m + 4n)² + (4m + 5n) (4m -5n)=16m²+25n²+40mn+25m²+16n²+40mn +16m²-25n²=57m²+16n²+80mn
(v) (.5p - 1.5q)²- (.5p - 1.5q)² +p²q²=3pq+p²q²=pq(3+pq)
(vi) (ab - bc)²+2ab²c= a²b²+b²c²-2ab²c+2ab²c=b²(ab²+cb²)
(vii) mb&sup4; + nb&sup4;

Question 4
Using identities, evaluate.
(a) 91²
(b) 89²
(c) 202²
(d) 999²
(e) 1.2²
(f) 397 x 403
(g) 48 x 52
(h) 5.1²
(i) 61²- 59²
(j) 11.1²- 9.9²
(k) 503 x 504
(l) 2.1 x 2.2
(m) 103 x 98
(n) 9.7 x 9.8
(o) 729²- 271²

Answer

(a) 91²=(90+1)²=8100+1+180=8281 (b) 89²=(90-1)²==8100+1-180=7921 (c) 202²=(200+2)²=40000+4+800=40804 (d) 999²=(1000-1)²=1000000+1 -2000=998001 (e) 1.2²=(1+.2)²=1+.04+.4=1.44 (f) 397 x 403=(400-3)(400+3)=160000-9=159991 (g) 48 x 52=(50-2)(50+2)=2500-4=2496 (h) 5.1²=(5+.1)²=25+.01+1=26.01 (i) 61²- 59²=(61-59)(61+59)=2*120=240 (j) 11.1²- 9.9²=(11.1-9.9)(11.1+9.9)=1.2*21=25.2 (k) 503 x 504=(500+3)(500+4)=250000+12+3500=253512 (l) 2.1 x 2.2=(2+.1)(2+.2)=4+.6+.02=4.62 (m) 103 x 98=(100+3)(100-2)=10000-6+100=10094 (n) 9.7 x 9.8=(10-.3)(10-.2)=100+.06-5=95.06 (o) 729²- 271²=(729-271)(729+271)=458*1000=458000

Question 5
Find the value of x if 8x=35² -27²

Answer

8x=35² -27²

8x=(35-27)(35+27)
8x=8*62
x=62

Question 6
(a) If a -1/a =4, find the value of a² + 1/a²
(b) If p +q =13 and pq =22, then p² + q²

Answer

(a) a- 1/a=4
Squaring both the sides
a² + 1/a²-2=16
a² + 1/a²=18

(b) p² + q²=(p+q)² -2pq
=13²-2*22=169-44=125

True and False

Question 7
(i) abc + bca + cab is a monomial.
(ii) An equation is true for all values of its variables.
(iii) $(a - b)^2 = a^2 - b^2$
(iv) a ( b + c) = ab + ac is a distributive Property
(v) $(a + b)^2 - (a - b)^2 = 2a^2 - 2b^2$

Answer

(i) False as the terms are same and it is equal to 3abc
(ii) False
(iii) false
(iv) True
(v) false

Fill in the blanks

Question 8
(i) $(a -2b))(a+ 2b)(a^2 + 4b^2)$ is _________
(ii) (a - b)²2 + ____________ = a² - b²
(iii) The coefficient in -7abc is _____
(iv)If a+b=9 and ab=14, the $a^2 + b^2$ is ______
(v) $82^2 -18^2$ is _____

Answer

(i) $(a -2b))(a+ 2b)(a^2 + 4b^2) = (a^2 -4b^2)(a^2 + 4b^2)=a^4 -16b^4)$
(ii) $2ab � 2b^2$ (iii) -7
(iv) 53 as $a^2 + b^2= (a+b)^2 -2 ab= 81 -28=53$
(v) $82^2 -18^2= (82+18)(82-18)=66 \times 100=6600$

Multiple Choice questions

Question 9
if $a^2 + b^2=74$ and $a+b=12$,then the value of ab is
(a) 35
(b) 32
(c) 40
(d) 37
Question 10
In a polynomial, the exponents of the variables are always
(a) integers
(b) positive integers
(c) non-negative integers
(d) non-positive integers
Question 11
if $x + \frac {1}{x}=4$ then the value $x - \frac {1}{x}$ is
(a) 16
(b) 4
(c) $\sqrt {12}$
(d) 0
Question 12
$(x^4 +1)(x^2 +1)(x+1)(x-1)$ is
(a) $x^4 -1$
(b) $x^4 +1$
(c) $x^{8} -1$
(d) $x^{16} -1$
Question 13
$x + \frac {1}{x}=5$ then $x^3 + \frac {1}{x^3}$ is
(a) 110
(b) 100
(c) 625
(d) 125

Answer

(9) (a) as $a^2 + b^2= (a+b)^2 -2 ab$ or $2ab=(a+b)^2 - a^2 + b^2= 70$ or ab=35
(10) (b)
(11) c
$(x - \frac {1}{x})^2= (x + \frac {1}{x})^2 -4= 16-4=12$
or $ (x - \frac {1}{x})=\sqrt {12}$
(12)(c) as $(x^4 +1)(x^2 +1)(x+1)(x-1)= (x^4 +1)(x^2 +1)(x^2 -1)= (x^4 +1)(x^4 -1)=x^{8} -1$
(13) (d)
$x + \frac {1}{x}=5$
$(x + \frac {1}{x})^3= 125$
$(x^3 + \frac {1}{x^3} + 3(x + \frac {1}{x})=125$
$x^3 + \frac {1}{x^3}=125 -15=110$

Match the column

Question 14

algebraic expressions and identities class 8 worksheets with solutions

Answer

(p) ->(iii)
(q) -> (i)
(r) -> (iv)
(s) -> (ii)

Summary

This algebraic expressions and identities class 8 worksheets with solutions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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