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

Factor the following:

(A) 9a2+b2+4c26ab+12ac4bc9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc
(B) 16s2+25t240st16s^2 + 25t^2 - 40st
(C) r2r42r^2 - r - 42
(D) 49g2+14gh+h249g^2 + 14gh + h^2
(E) 64u2+121v2+4w2176uv32uw+44vw64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw                                         [Page No. 82]

Answer: Verified

11. (A) 9a2+b2+4c26ab+12ac4bc=(3a)2+(b)2+(2c)2+2(3a)(b)+2(b)(2c)+2(3a)(2c)=(3ab+2c)2(B)16s2+25t240st=(4s)22(4s)(5t)+(5t)2=(4s5t)2(C)Use a+b=1,ab=42So, a=7,b=6r2r42=(r7)(r+6)(D)49g2+14gh+h2=(7g)2+2(7g)(h)+h2=(7g+h)2(E)64u2+121v2+4w2176uv32uw+44vw=(8u)2+(11v)2+(2w)2+2(8u)(11v)+2(11v)(2w)+2(8u)(2w)=(8u11v2w)2\begin{aligned} 11. \text{ (A) } 9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc \\ &= (3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b) \\ &\quad (2c) + 2(3a)(2c) \\ &= (3a - b + 2c)^2 \\ (B) \quad 16s^2 + 25t^2 - 40st &= (4s)^2 - 2(4s)(5t) + (5t)^2 \\ &= (4s - 5t)^2 \\ (C) \quad \text{Use } a + b = -1, ab = -42 \\ \text{So, } a = -7, b = 6 \\ r^2 - r - 42 &= (r - 7)(r + 6) \\ (D) \quad 49g^2 + 14gh + h^2 &= (7g)^2 + 2(7g)(h) + h^2 \\ &= (7g + h)^2 \\ (E) \quad 64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw \\ &= (8u)^2 + (-11v)^2 + (-2w)^2 + 2(8u)(-11v) \\ &\quad + 2(-11v)(-2w) + 2(8u)(-2w) \\ &= (8u - 11v - 2w)^2 \end{aligned}

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