Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.3
(2) given f(x) = x3 – px2 + 6x – p ; x – p
∴ using remainder theorem, we can say that,
Remainder of f(x) = f(p)
∴ f(p) = p3 – p x p2 + 6 x p – p
= p3 – p3 + 6p – p
= 5p
(5) Given, f(x) 2x3 + ax2 + 3x – 5 ; p(x)x3 + x2 – 4x + a
Divided by x -2
By using remainder theorem we get
∴ f(2) = 2 x 23 + a x 22 + 3 x 2 – 5
= 16 + 4a + 6 – 5
= 4a + 17
P(2) = 23 + 22 – 4×2 + a
= 8 + 4 – 8 + a
= a + 4
Now,
Given that remainder of f(x) and p(x) is same when divided by x – 2
∴ f(2) = p(2)
Or, 4a + 17 = a + 4
Or, 4a – a = 4 – 17
Or, 3a = -13
Or, a = -13/3
(6) given, f(x) = x3 + ax2 + 5 ; p(x) = x3 -2x2 + a
Divided by x + 2
∴ using remainder theorem,
f(-2) = (-2)3 + a x (-2)2 + 5
= 8 + a x 4 + 5
= 4a – 3
P(-2) = (-2)3 – 2(-2)2 + a
= -8 – 8 + a
= a – 16
Now, given that
f(-2) = p(-2)
∴ 4a – 3 = a – 16
Or, 4a – a = -16 + 3
Or, 3a = -13
Or, a = -13/3
I 9th bit is missing
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state syllabus answers are difficult to find where cbse is very easy to find
9th bit is missing