**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 = x^{82}

An example of a binomial having a degree of 99 = x^{99 }+ x

**2. Compute the value of 9x ^{2} + 4y^{2} if xy = 6 and 3x + 2y = 12.**

**Solution:**

Consider the equation 3x + 2y = 12

Now, square both sides:

(3x + 2y)^{2} = 12^{2}

=> 9x^{2} + 12xy + 4y^{2} = 144

=>9x^{2} + 4y^{2} = 144 – 12xy

From the questions, xy = 6

So,

9x^{2} + 4y^{2} = 144 – 72

Thus, the value of 9x^{2} + 4y^{2 }= 72

**3. ** Find the value of the polynomial 5x – 4x^{2} + 3 at x = 2 and x = –1

**Solution:**

Let the polynomial be f(x) = 5x – 4x^{2} + 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 – 4x^{2} + 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 – 4x^{2} + 3 at x = -1 is -6.

**4. Calculate the perimeter of a rectangle whose area is 25x**^{2}** – 35x + 12. **

**Solution:**

Given,

Area of rectangle = 25x^{2} – 35x + 12

We know, area of rectangle = length × breadth

So, by factoring 25x^{2} – 35x + 12, the length and breadth can be obtained.

25x^{2} – 35x + 12 = 25x^{2 }– 15x – 20x + 12

=> 25x^{2} – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x^{2} – 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 x**^{3}** + y**^{3}** + z**^{3}** – 3xyz if x**^{2}** + y**^{2}** + z**^{2}** = 83 and x + y + z = 15**

**Solution:**

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)^{2} = a^{2} + b^{2} + c^{2 }+ 2(ab + bc + ca)

So,

(x + y + z)^{2} = x^{2 }+ y^{2} + z^{2} + 2(xy + yz + xz)

From the question, x^{2} + y^{2} + z^{2} = 83 and x + y + z = 15

So,

15^{2} = 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),

x^{3} + y^{3} + z^{3} – 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, x^{3} + y^{3} + z^{3} – 3xyz = 15(83 – 71)

=> x^{3} + y^{3} + z^{3} – 3xyz = 15 × 12

Or, x^{3} + y^{3} + z^{3} – 3xyz = 180