Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

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).
Solution: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}
Solution:(i)(ii)

Question 9.Using factor theorem, factorize each of the following polynomials: x³ + 6x² + 11x + 6
Solution: 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)