Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

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Q 1. Show that (x + 3) is a factor of x³ + x² – 4x + 6.
Sol: ∵ p(x) = x³ + x² – 4x + 6
Since (x + 3) is a factor, then x + 3 = 0 ⇒ x = – 3
∴ p( – 3) = ( – 3)³ + ( – 3)²  – 4( – 3) + 6
= – 27 + 9 + 12 + 6
= 0
∴ (x + 3) is a factor of x³ + x² – 4x + 6

Q 2.Define zero or root of a polynomial
Sol: Zero or root, is a solution to the polynomial equation, f(y) = 0.
It is that value of y that makes the polynomial equal to zero

Q 3. Find the value of a such that (x + α ) is a factor of the polynomial
f(x) = x⁴ – α 2x² + 2x + α + 3.
Sol: Here f(x) = x⁴  – α² x² + 2x + α + 3
Since, (x + a) is a factor of f(x)
∴ f ( – α) = 0
⇒ ( – α)⁴ – α² ( – α)² + 2( – α) + a + 3 = 0
⇒ α⁴ – α⁴ – 2α. α+ 3= 0
⇒ – α + 3 = 0
⇒ α= 3

Q 4.Find the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.
Sol: Let the polynomial be f(x) = 5x – 4x² + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)² + 3
⇒  f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is -3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)² + 3
⇒ f(–1) = –5 –4 + 3 = -6
The value of the polynomial 5x – 4x² + 3 at x = -1 is -6.

Q5. Factorise x² + 1/x² + 2 – 2x – 2/x.
Sol:  x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)

= (x + 1/x)² – 2(x + 1/x)

= (x + 1/x)(x + 1/x – 2).

Q6. Show that (x – 5) is a factor of: x³ – 3x²  – 13x + 15
Sol: ∵ p(x) = x³ – 3x² – 13x + 15
Since (x – 5) is a factor, then x – 5 = 0
⇒ x = 5
∴ p(5) = (5)³  – 3(5)²  – 13(5) + 15
= 125 – 75 – 65 + 15 = 140 – 140 = 0
∴ (x – 5) is a factor of x³  – 3x² – 13x + 15.

Question 7. Show that x³ + y³ = (x + y )(x² – xy + y²).
Sol: Since (x + y)³= x³+ y³ + 3xy(x + y)
∴ x³+ y³= [(x + y)³] –3xy(x + y)
= [(x + y)(x + y)²] – 3xy(x + y)
= (x + y)[(x + y)²– 3xy]
= (x + y)[(x² + y²+ 2xy) – 3xy]
= (x + y)[x²+ y²– xy]
= (x + y)[x²+ y²– xy]
Thus, x³+ y³= (x + y)(x²– xy + y²)

Question 8.Evaluate each of the following using identities:
(i) (399)^{2
(ii) (0.98)2}
Sol: (i)
3992 = (400 − 1)2
= (400)2 + (1)2 − 2 × 400 × 1
[Use identity: (a − b)2 = a2 + b2 − 2ab]
Here, a = 400 and b = 1
= 160000 + 1 − 800
= 159201
So, 3992 = 159201

(ii) 
(0.98)2 = (1 − 0.02)2
[Use identity: (a − b)2 = a2 + b2 − 2ab]
= (1)2 + (0.02)2 − 2 × 1 × 0.02
= 1 + 0.0004 − 0.04
= 1.0004 − 0.04
= 0.9604
So, (0.98)2 = 0.9604

Question 9.Using factor theorem, factorize each of the following polynomials: x³ + 6x² + 11x + 6
Sol: Let f(x) = x³ + 6x² + 11x + 6
Step 1: Find the factors of constant term
Here constant term = 6
Factors of 6 are ±1, ±2, ±3, ±6
Step 2: Find the factors of f(x)
Let x + 1 = 0
⇒ x = -1
Put the value of x in f(x)
f(-1) = (−1)³ + 6(−1)² + 11(−1) + 6
= -1 + 6 -11 + 6
= 12 – 12
= 0
So, (x + 1) is the factor of f(x)
Let x + 2 = 0
⇒ x = -2
Put the value of x in f(x)
f(-2) = (−2)³ + 6(−2)² + 11(−2) + 6 = -8 + 24 – 22 + 6 = 0
So, (x + 2) is the factor of f(x)
Let x + 3 = 0
⇒ x = -3
Put the value of x in f(x)
f(-3) = (−3)³ + 6(−3)² + 11(−3) + 6 = -27 + 54 – 33 + 6 = 0
So, (x + 3) is the factor of f(x)
Hence, f(x) = (x + 1)(x + 2)(x + 3)