Important Polynomials Questions For Class 9- Chapter 2 (With Solutions)
1. Give an example of a monomial and a binomial having degrees as 82 and 99, respectively.
Solution:
An example of monomial having a degree of 82 = x82
An example of a binomial having a degree of 99 = x99 + x
2. Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12.
Solution:
Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)2 = 122
=> 9x2 + 12xy + 4y2 = 144
=>9x2 + 4y2 = 144 – 12xy
From the questions, xy = 6
So,
9x2 + 4y2 = 144 – 72
Thus, the value of 9x2 + 4y2 = 72
3. Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1
Solution:
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.
4. Calculate the perimeter of a rectangle whose area is 25x2 – 35x + 12.
Solution:
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
5. Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15
Solution:
Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
So,
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz)
From the question, x2 + y2 + z2 = 83 and x + y + z = 15
So,
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x3 + y3 + z3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
Now,
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x3 + y3 + z3 – 3xyz = 15(83 – 71)
=> x3 + y3 + z3 – 3xyz = 15 × 12
Or, x3 + y3 + z3 – 3xyz = 180