Q1. Write the numerical coefficient and degree of each term of:
Q2. Find the remainder when x^{3} – ax^{2} + 4x – a is divided by (x – a).
p(x) = x^{3 }– ax^{2} + 4x – a
(x – a) = 0
⇒ x = a
∴ p(a) = (a)^{3} – a(a)^{2} + 4(a) – a = a^{3} – a^{3 }+ 4a – a = 4a – a = 3a
∴ The required remainder = 3a.
Q3. When the polynomial kx^{3} + 9x^{2} + 4x – 8 is divided x + 3, then a remainder 7 is obtained.
Find the value of k.
Here, p(x) = kx^{3} + 9x^{2 }+ 4x – 8
Since, Divisor = x + 3
∴ x + 3 = 0
⇒ x = –3
∴ p(–3) = 7
⇒ k(–3)^{3} + 9(–3)^{2 }+ 4(–3) – 8 = 7
⇒ –27k + 81 – 12 – 8 = 7
⇒ –27k = 7 – 81 + 12 + 8
⇒ –27k = 27 – 81
⇒ –27k = –54
⇒ k= (54/27) = 2
Thus, the required value of k = 2.
Q4. (a) For what value of k, the polynomial x^{2} + (4 – k)x + 2 is divisible by x – 2?
(b) For what value of ‘m’ is x^{3} – 2mx^{2} + 16 is divisible by (x + 2)?
(a) Here p(x) = x^{2 }+ 4x – kx + 2
If p(x) is exactly divisible by x – 2, then p(2) = 0
i.e. (2)^{2} + 4(2) – k(2) + 2 = 0
⇒ 4 + 8 – 2k + 2 = 0
⇒ 14 – 2k = 0
⇒ 2k = 14
⇒ k= 14/2 = 7
Thus, the required value of k is 7.(b) Here, p(x) = x^{3} – 2mx^{2} + 16
∴ p(–2) = (–2)^{3} –2(–2)^{2}m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1
Q5. Factorize x^{2} – x – 12.
We have x^{2} – x – 12
⇒ x^{2 }– 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x^{2 }– x – 12 = (x – 4)(x + 3)
Q6. If x + (1/2x) = 5, then find the value of x^{2} +
We have x + (1/2x) = 5
Squaring both sides, we get:
⇒
⇒
⇒
Thus, the required value of
Q7. Check whether (x – 1) is a factor of the polynomial x^{3} – 27x^{2} + 8x + 18.
Here, p(x) = x^{3 }– 27x^{2} + 8x + 18, (x – 1) will be a factor of p(x) only if (x – 1) divides p(x) leaving a remainder 0.
For x – 1 = 0
⇒ x = 1
∴ p(1) = (1)^{3} – 27(1)^{2} + 8(1) + 18
⇒ 1 – 27 + 8 + 18
⇒ 27 – 27
⇒ 0
Since, p(1) = 0
∴ (x – 1) is a factor of p(x).
Thus, (x – 1) is a factor of x^{3} – 27x^{2} + 8x + 18.
Q8. Find the remainder when f(x) =
Here f(x) =
Divisor = x+(2/3)
since, x+(2/3) = 0 ⇒ x = 2/3
∴ Remainder = f (2/3)
i.e. Remainder
Thus, the required remainder = (10/27)
Q9. Find the value of k, if (x – k) is a factor of x^{6} – kx^{5} + x^{4} – kx^{3} + 3x – k + 4.
Here, p(x) = x^{6} – kx^{5} + x^{4 }– kx^{3} + 3x – k + 4
If (x – k) is a factor of p(x), then p(k) = 0
i.e (k)^{6} – k(k^{5}) + k^{4} – k(k^{3}) + 3k – k + 4 = 0
⇒ k^{6 }– k^{6} + k^{4} – k^{4} + 3k – k + 4 = 0
⇒ 2k + 4 = 0
⇒ 2k = – 4
⇒ k = (4/2) = –2
Thus, the required value of k is –2.
Q10. Factorize: 9a^{2} – 9b^{2} + 6a + 1
9a^{2} – 9b^{2} + 6a + 1
⇒ [9a^{2} + 6a + 1] – 9b^{2 }
⇒ [(3a)^{2} + 2(3a)(1) + (1)2] – (3b)^{2 }
⇒ (3a + 1)^{2} – (3b)^{2}
⇒ [(3a + 1) + 3b][(3a + 1) – 3b] {using x^{2 }– y^{2} = (x – y)(x + y)}
⇒ (3a + 1 + 3b)(3a + 1 – 3b)
Q11. Find the value of x^{3} + y^{3} – 12xy + 64, when x + y = –4.
x^{3} + y^{3 }– 12xy + 64
⇒ (x)^{3} + (y)^{3} + (4)^{3 }– 3(x)(y)(4)
⇒ [x^{2} + y^{2} + 42 – xy – y. 4 – 4 . x](x + y + 4)
⇒ [x^{2} + y^{2} + 16 – xy – 4y – 4x][x + y + 4] ...(1)
Since, x + y = –4 ∴ x + y + 4 = 0 ...(2)
From (1) and (2), we have x^{3} + y^{3 }– 12xy + 64
⇒ [x^{2 }+ y^{2} + 16 – xy – 4y – 4x][0] = 0
Thus, x^{3} + y^{3} – 12xy + 64 = 0.
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ValueBased Questions: Polynomials Doc  1 pages 
Chapter Notes: Polynomials Doc  4 pages 
NCERT Solutions: Polynomials (Exercise 2.1) Doc  4 pages 
1. What is a polynomial? 
2. How do you identify the degree of a polynomial? 
3. What is the leading coefficient of a polynomial? 
4. Can a polynomial have a negative degree? 
5. What is the difference between a monomial and a polynomial? 
48 videos387 docs65 tests

ValueBased Questions: Polynomials Doc  1 pages 
Chapter Notes: Polynomials Doc  4 pages 
NCERT Solutions: Polynomials (Exercise 2.1) Doc  4 pages 

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