Chapter 4
Exploring Algebraic Identities
CBSE Class 9
Mathematics Solutions
Educart Mathematics class Class 9 NCERT Exemplar cover
Question:

Factor completely:

(A) 9x2+24xy+16y29x^2 + 24xy + 16y^2

(B) 4s2+20st+25t24s^2 + 20st + 25t^2

(C) 49x2+28xy+4y249x^2 + 28xy + 4y^2

(D) 64p2+323pq+49q264p^2 + \frac{32}{3}pq + \frac{4}{9}q^2

* (E) 3a2+4ab+43b23a^2 + 4ab + \frac{4}{3}b^2

* (F) 95s2+6sv+5v2\frac{9}{5}s^2 + 6sv + 5v^2                                     [Page No. 74]

Answer: Verified

(A) (3x)2+2(3x)(4y)+(4y)2=(3x+4y)2(3x)^2 + 2(3x)(4y) + (4y)^2 = (3x + 4y)^2

(B) (2s)2+2(2s)(5t)+(5t)2=(2s+5t)2(2s)^2 + 2(2s)(5t) + (5t)^2 = (2s + 5t)^2

(C) (7x)2+2(7x)(2y)+(2y)2=(7x+2y)2(7x)^2 + 2(7x)(2y) + (2y)^2 = (7x + 2y)^2

(D) (8p)2+2(8p)(2q3)+(2q3)2=(8p+2q3)2(8p)^2 + 2(8p)\left(\frac{2q}{3}\right) + \left(\frac{2q}{3}\right)^2 = \left(8p + \frac{2q}{3}\right)^2

(E) Take 13\frac{1}{3} common from given expression:

3a2+4ab+43b23a^2 + 4ab + \frac{4}{3}b^2

=13[9a2+12ab+4b2]= \frac{1}{3}[9a^2 + 12ab + 4b^2]

=13[(3a)2+2(3a)(2b)+(2b)2]= \frac{1}{3}[(3a)^2 + 2(3a)(2b) + (2b)^2]

=(13)(3a+2b)2= \left(\frac{1}{3}\right)(3a + 2b)^2

(F) Take 15\frac{1}{5} common from given expression:

=(15)[9s2+30sv+25v2]= \left(\frac{1}{5}\right)[9s^2 + 30sv + 25v^2]

=(15)[(3s)2+2(3s)(5v)+(5v)2]= \left(\frac{1}{5}\right)[(3s)^2 + 2(3s)(5v) + (5v)^2]

=(15)(3s+5v)2= \left(\frac{1}{5}\right)(3s + 5v)^2

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