Previous Year Questions : Polynomials | Mathematics (Maths) Class 9 PDF Download

Sol:

(i) x²+4

• Quadratic Polynomial (degree 2)

(ii) 2x−3x²

• Quadratic Polynomial (degree 2)

(iii) 7y³−5y²+4

• Cubic Polynomial (degree 3)

(iv) 3x+2

• Linear Polynomial (degree 1)

(v) 5x²−6x+8

• Quadratic Polynomial (degree 2)

Sol:

(i) Find p(−1):

Substitute x=−1 into the polynomial p(x) = x³−4x+2:

(ii) Find q(0):

Substitute y=0 into the polynomial q(y) = 4y²−3y +7:

Sol:

∵ (x + y)³ = x³ + y³ + 3xy (x + y)
∴ [x³ + y³ + 3xy (x + y)] –  [3xy (x + y)] = x³ + y³ 
⇒ (x + y)³ – 3xy(x + y) = x³ + y³
Thus, value of x³ + y³ is (x + y)³ – 3xy (x + y) = x³ + y³

Sol:Let p(x)=x²−2x−3.

Now, check the values of the polynomial at x=−1 and x=3:

For x=−1:

p(−1)=(−1)²−2(−1)−3 =1+2−3 =0

So, −1 is a zero of the polynomial.

For x=3:

p(3)=3²−2(3)−3 =9−6−3 =0

So, 3 is also a zero of the polynomial.

Conclusion: Both -1 and 3 are zeros of the polynomial x²−2x−3.

Sol:The degree of 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is 4.

Sol:∵ p(x) = 0 ⇒ 2x + 5 = 0

⇒ x = -5/2 

∴ zero of 2x + 5 is -5/2 

Sol:∵ 2x² + 7x – 4 = 2x² + 8x – x – 4
⇒ 2x (x + 4) – 1 (x + 4) = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x = – 4, x = 1/2,
∴ One of the zero of 2x² + 7x – 4 is 1/2.

Sol:Let p(x)=x²+3x−4.

For x=1:

So, x=1 is a zero of the polynomial.

For x=−1:

So, x=−1 is not a zero of the polynomial.

Sol:As we know, degree is the highest power in the polynomial

(i) Degree of the polynomial 7x³ + 4x² – 3x + 12 is 3

(ii) Degree of the polynomial 12 – x + 2x³ is 3

Sol:Given expression is $x^\{100\}\; -\; 1$x¹⁰⁰−1.

Since all the exponents of $x$x are whole numbers and the highest exponent of $x$x is 100, it is a polynomial of degree 100.

Sol:(a + b + c)² + (a − b + c)²

= (a² + b² + c² + 2ab+2bc+2ca) + (a² + (−b)² + c² −2ab−2bc+2ca)
= 2a² + 2 b² + 2c² + 4ca

Sol:

Sol:

We have  x² – x – 12
⇒ x² – 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x² – x – 12 = (x – 4)(x + 3)

Sol:

Using identities:

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

(i) (a + 2b + c)²

= a² + (2b)² + c² + 2a(2b) + 2ac + 2(2b)c

= a² + 4b² + c² + 4ab + 2ac + 4bc

(ii) (2a − 3b − c)²

= [(2a) + (−3b) + (−c)]²

= (2a)² + (−3b)² + (−c)² + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c)

= 4a² + 9b² + c² − 12ab + 6bc − 4ca

Sol:

A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents.is not a polynomial.

Sol:

Here, p(x) = x³ – 27x² + 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 x³ – 27x² + 8x + 18.

Sol:

(a) Here p(x) = x² + 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) = x³ – 2mx² + 16
∴ p(–2) = (–2)³ –2(–2)²m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1

Sol:f(x) = 2x³ – 13x² + 17x + 12

(i) f(2) = 2(2)³ – 13(2)² + 17(2) + 12

= 2 x 8 – 13 x 4 + 17 x 2 + 12

= 16 – 52 + 34 + 12

= 62 – 52

= 10

(ii) f(-3) = 2(-3)³ – 13(-3)² + 17 x (-3) + 12

= 2 x (-27) – 13 x 9 + 17 x (-3) + 12

= -54 – 117 -51 + 12

= -222 + 12

= -210

(iii) f(0) = 2 x (0)³ – 13(0)² + 17 x 0 + 12

= 0-0 + 0+ 12

= 12

Sol:

1. a³ + b³ = (a + b)(a² + b² – ab)

2. a³ – b³ = (a – b)(a² + b² + ab)

3. (a + b)(a-b) = a² – b²

4. (a + b)² = a² + b² + 2ab and

5. (a – b)² = a² + b² – 2ab ]

(i) (x + 3)³ + (x – 3)³

Here a = (x + 3), b = (x – 3)(ii) (x/2 + y/3)³ – (x/2 – y/3)³

Here a = (x/2 + y/3) and b = (x/2 – y/3)

(iii) (x + 2/x)³ + (x – 2/x)³

Here a = (x + 2/x) and b = (x – 2/x)