Next exercise – Polynomials – Exercise 8.2 – Class X
Polynomials – Exercise 8.3
1.In each of the following cases, use the remainder theorem and find the remainder when p(x) is divided by g(x)
(i) p(x) = x3 + 3x2 – 5x + 8 ; g(x) = x – 3
(ii) p(x) = 4x3 – 10x2 + 12x – 3 ; g(x) = x + 1
(iii) p(x) = 2x4 – 5x2 + 15x – 6 ; g(x) = x – 2
(iv) p(x) = 4x3 – 12x2 + 14x – 3 ; g(x) = 2x – 1
(v) p(x) = 7x3 – x2 + 2x – 1 ; g(x) = 1 – 2x
- If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 4x – a) leave the same remainder when divided by (x – 1), find the value of a.
- The polynomials (2x3 – 5x2 + x + a) and (ax3 + 2x2 – 3)when divided by (x – 2) leave the remainders R1 and R2 respectively. Find the value of a in each of the following cases, if
(i) R1 = R1
(ii)2R1 + R2 = 0
(iii) R1 – 2R2 = 0
Polynomials – Exercise 8.3 – Solutions:
- In each of the following cases, use the remainder theorem and find the remainder when p(x) is divided by g(x)
(i) p(x) = x3 + 3x2 – 5x + 8 ; g(x) = x – 3
Solution:
p(x) = x3 + 3x2 – 5x + 8
g(x) = x – 3
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = x3 + 3x2 – 5x + 8 is p(3)
p(3) = (3)3 + 3(3)2 – 5(3) + 8
= 27 + 27 – 15 + 8
= 47
Therefore, the remainder r(x) = 47.
(ii) p(x) = 4x3 – 10x2 + 12x – 3 ; g(x) = x + 1
Solution:
p(x) = 4x3 – 10x2 + 12x – 3
g(x) = x + 1
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = 4x3 – 10x2 + 12x – 3 is p(-1)
p(-1) = 4(-1)3 – 10(-1)2 + 12(-1) – 3
= 4(-1) – 10(1) + 12 (-1) – 3
= -4 – 10 -12 – 3
= -29
Therefore, the remainder r(x) = -29.
(iii) p(x) = 2x4 – 5x2 + 15x – 6 ; g(x) = x – 2
Solution:
p(x) = 2x4 – 5x2 + 15x – 6
g(x) = x – 2
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = 2x4 – 5x2 + 15x – 6 is p(2)
p(2) = (2)3 + 3(2)2 – 5(2) + 8
= 8 + 12 – 10 + 8
= 18
Therefore, the remainder r(x) = 18.
(iv) p(x) = 4x3 – 12x2 + 14x – 3 ; g(x) = 2x – 1
Solution:
p(x) = 4x3 – 12x2 + 14x – 3
g(x) = 2x – 1
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = 4x3 – 12x2 + 14x – 3 is p(1/2)
p(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2) – 3
= 4x1/8 – 12 x 1/4 + 14x1/2 – 3
= 1/2 – 3 + 7 – 3
=3/2
Therefore, the remainder r(x) = 3/2
(v) p(x) = 7x3 – x2 + 2x – 1 ; g(x) = 1 – 2x
Solution:
p(x) = 7x3 – x2 + 2x – 1
g(x) = 1 – 2x
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = 7x3 – x2 + 2x – 1 is p(-1/2)
p(-1/2)= 7(-1/2)3 – (-1/2)2 + 2(-1/2) – 1
= -7x1/8 – 1/4 – 1 – 1
= -7 – 2 – 8 – 8/8
= –25/8
Therefore, the remainder r(x) = –25/8
- If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 4x – a) leave the same remainder when divided by (x – 1), find the value of a.
Solution:
Let p(x) = (2x3 + ax2 + 3x – 5) and g(x) = (x3 + x2 – 4x – a)
Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).
Therefore, by the remainder theorem, the remainder of p(x) = 2x3 + ax2 + 3x – 5 and g(x) = (x3 + x2 – 4x – a) is p(1) and g(1)
By the given condition, p(1) = g(1)
p(1) = 2(1)3 + a(1)2 + 3(1) – 5 = 2 + a + 3 – 5 = a
g(1) = (1)3 + (1)2 – 4(1) – a = 1 + 1 – 4 – a = -2 – a
Since p(1) = g(1)
⇒ a = -2 – a
a + a = – 2
2a = -2
a = – 1
- The polynomials (2x3 – 5x2 + x + a) and (ax3 + 2x2 – 3)when divided by (x – 2) leave the remainders R1 and R2 respectively. Find the value of a in each of the following cases, if
(i) R1 = R1
(ii)2R1 + R2 = 0
(iii) R1 – 2R2 = 0
Solution:
Let p(x) = (2x3 – 5x2 + x + a) and g(x) = (ax3 + 2x2 – 3)
R1 is the remainder when p(x) = (2x3 – 5x2 + x + a)is divided by (x – 2).
⸫R1 = p(2)
p(2) = (2(2)3 – 5(2)2 + (2) + a) = 16 – 20 + 2 + a = -2 + a
⇒ R1 = a – 2
R2 is the remainder when g(x) = (ax3 + 2x2 – 3)is divided by (x – 2).
R2 = g(2) = a(2)3 + 2(2)2 – 3 = 8a + 8 – 3 = 8a + 5
⇒ R2 = 8a + 5
(i) R1 = R2
a – 2 = 8a + 5
a – 8a = 5 + 2
-7a = 7
a = -1
(ii)2R1 + R2 = 0
2(a – 2) + (8a + 5) = 0
2a – 4 + 8a + 5 = 0
10 a + 1 = 0
10 a = -1
a = –1/10
(iii) R1 – 2R2 = 0
(a – 2) – 2(8a + 5) = 0
a – 2 – 16a – 10 = 0
– 15a – 12 = 0
a = – 12/15 = –4/5
Next Exercise – Polynomials 8.4 – Exercise 8.4 – Class X
