Class 9 Maths Question Answers - Polynomials

Q1: Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Sol:An example of a monomial having a degree of 82 = x⁸²
An example of a binomial having a degree of 99 = x⁹⁹ + x

Q2: 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.

Q3:Factorise the quadratic polynomial by splitting the middle term: x² + 14x + 45
Sol: x² + 14x + 45

here a = 1, b = 14, c = 45

ac = 45 = 9 x 5 , b = 14 = 9 + 5

x² + 14x + 45

= x² + 9x + 5x + 45

= x( x + 9 ) + 5(x + 9)

= (x + 9) (x + 5) 

Q4: Check whether 3 and -1 are zeros of the polynomial $x\; +\; 4$x+4.

Sol: Let p(x)=x+4.

Now, check for each value:

p(3) =3+4 =7

$p(-1)\; =\; -1\; +\; 4\; =\; 3$p(−1)=−1+4 =3

Therefore, 3 and -1 are not zeros of the polynomial $x+4x\; +\; 4$x+4. 

Q5: Check whether (7 + 3x) is a factor of (3x³ + 7x).
Sol:Let p(x) = 3x³ + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, we know that when p(x) is divided by g(x) then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)³ + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).

Q6: Verify whether 1 and -2 are zeros of the polynomial $x^2\; +\; 3x$x²+3x.
Sol: Let p(x)=x2+3x.

Now, check for each value:

• p(1) =12+3(1)=1+3 =4
• p(−2) =(−2)2+3(−2)=4−6 =−2

Therefore, 1 and -2 are not zeros of the polynomial x²+3x.

Q7: Calculate the perimeter of a rectangle whose area is 25x² – 35x + 12. 
Sol:Given,
Area of rectangle = 25x² – 35x + 12
We know, area of rectangle = length × breadth
So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.
25x² – 35x + 12 = 25x² – 15x – 20x + 12
⇒ 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
⇒ 25x² – 35x + 12 = (5x – 3)(5x – 4)
So, the length and breadth are (5x – 3)(5x – 4).
Now, perimeter = 2(length + breadth)
So, perimeter of the rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14
So, the perimeter = 20x – 14

Q8: Find the value of k, if $x\; +\; 2$x+2 is a factor of 3x³+5x²−2x+k.
Sol: As $x\; +\; 2$x+2 is a factor of $p(x)\; =\; 3x^3\; +\; 5x^2\; -\; 2x\; +\; k$p(x)=3x³+5x²−2x+k, we know that p(−2)=0.

Now, calculate p(−2) :

$p(-2)\; =\; 3(-2)^3\; +\; 5(-2)^2\; -\; 2(-2)\; +\; k$p(−2) =3(−2)3+5(−2)2−2(−2)+k

p(−2) =3(−8)+5(4)+4+k = −24+20+4+k

$0\; =\; -24\; +\; 20\; +\; 4\; +\; k$0 =−24+20+4+k

$0\; =\; 0\; +\; k$0 = 0+k

k =0. 

Q9: Find the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.
Sol:Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)³ + a(2)² + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)³ + a(½)² + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.

Q10: 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).