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#### Class ten maths Polynomial NCERT Solutions

## Chapter 2: Polynomial NCERT Solutions

**TEXTBOOK’S EXERCISE – 2.2 **

**2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

**Sol. **We know that a quadratic polynomial with zeroes α and β is given by

∴ A required quadratic polynomial is

∴ A required quadratic polynomial is

∴ A required quadratic polynomial is x^{2}+ √5.

∴ A required quadratic polynomial is x^{2} – x + 1.

∴ A required quadratic polynomial is

∴ A required quadratic polynomial is x^{2} – 4x + 1

**TEXTBOOK’S EXERCISE – 2.3 **

**1. **Divide the polynomial **p(x) **by the polynomial **g(x) **and find the quotient and remainder in each of the following :

- P(x) = x
^{3}– 3x^{2}+ 5x -3

g(x) = x^{2}-2

- P(x) =x
^{4}– 3x^{2}+ 4x -3

g(x) = x^{2}+1-x

- P(x) = x
^{4}– 5x +6

g(x) = 2 – x^{2}

∴ By division algorithm,

x^{3}– 3x^{2} + 5x -3= (x-3)(x^{2}-2)+(7x-9)

Here, quotient = x – 3 and remainder = 7x – 9

By division algorithm,

By division algorithm,

x^{4} – 5x^{2}
+ 6 = (–x^{2} – 2)(–x^{2} + 2) + (–5x + 10)

Here, quotient = − x^{2} −2

and remainder = – 5x + 10

**2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :**

Since, remainder is zero, therefore

t^{2} – 3 is factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

Since, remainder is 0, therefore, x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Since, remainder is
not zero, therefore, x^{3}
– 3x + 1 is not a factor of x5 – 4x^{3} + x^{2}
+ 3x + 1.

**3. Obtain all other zeroes of 3x**^{4}** + 6x**^{3}** – 2x**^{2}** – 10x – 5, if two of its zeroes are**

**Sol. **Since two zeroes are

factors of the given polynomial.

is a factor of the given polynomial. Now, let us divide the given polynomial by

No, factorizing

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