## MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.

Find the value of the polynomial 5x – 4x^{2} + 3 at

- x = 0
- x = – 1
- x = 2.

Solution:

Let f(x) = 5x – 4x^{2} + 3

1. Value of f(x) at x = 0

= f(0) = 5(0) – 4(0)^{2} + 3 = 3

2. Value of f(x) at x = – 1

= f(- 1) = 5(- 1) – 4(- 1)^{2} + 3

= – 5 – 4 + 3 = – 6

3. Value of f(x) at x = 2

= f(2) = 5(2) – 4(2)^{2} + 3

= 10 – 16 + 3 = – 3.

Question 2.

Find p(0), p(1) and p(2) for each of the following polynomials:

- p(y) = y
^{2}– y + 1 - p(t) = 2 + t + 2t
^{2}– t^{3} - p(x) = x
^{3} - p(x) = (x – 1)(x + 1)

Solution:

1. p(y) = y^{2} – y + 1

P(0) = (0)2 – (0) + 1

= p(1) = (1)^{2} – (1) + 1 = 1,

and p(2)^{2} = (2)^{2} – (2) + 1

= 4 – 2 + 1 = 3.

2. p(t) = 2 + t + 2t^{2} – t^{3}

p(0) = 2 + 0 + 2(0)^{2} – (0)^{2} = 2

p(1) = 2 + 1 + 2(1)^{2} – (1)^{2}

= 2 + 1 + 2 – 1 = 4

and p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

= 2 + 2 + 8 – 8 = 4

3. p(x) = x^{3}

p(0) = (0)^{2} = 0,

p(1) = (1)^{2} = 1,

and p(2) = (2)^{3} = 8.

4. p(x) = (x – 1) (x + 1)

p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1

P(1) = (1 – 1) (1 + 1) = (0) (2) = 0 and

p(2) = (2 – 1) (2 + 1) = (1) (3) = 3

Question 3.

Verify whether the following are zero of the polynomial, written against them.

- p(x) = 3x + 1, x = – \(\frac{1}{3}\)
- p(x) = 5x – π, x = \(\frac{4}{5}\)
- p(x) = x2 -1, x = 1, – 1
- p(x) = (x + 1) (x – 2), x – 1, 2
- p(x) = x
^{2}, x = 0 - p(x) = lx + m, x = – \(\frac{m}{l}\)
- p(x) = 3x
^{2}– 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\) - p(x) = 2x + 1, x = \(\frac{1}{2}\)

Solution:

1. p(x) = 3x + 1

x = – \(\frac{1}{3}\)

p(- \(\frac{1}{3}\)) = 3(- \(\frac{1}{3}\)) + 1 = 0

= – 3 x \(\frac{1}{3}\) + 1 = – 1 + 1 = 0

As p(- \(\frac{1}{3}\)) = 0, – \(\frac{1}{3}\) is the zero of p(x).

2. p(x) = 5x – π

x = \(\frac{4}{5}\)

p(\(\frac{4}{5}\)) = 5 x \(\frac{4}{5}\) – π = 4 – π

As p(\(\frac{4}{5}\)) ≠ 0,

\(\frac{4}{5}\) is not the zero of p(x).

3 p(x) = x^{2} – 1, x = 1, -1

P(1) = (1)^{2} – 1 = 1 – 1 = 0

p(-1) = (-1)^{2} – 1 = 1 – 1 = 0

As p(1) = 0 and p(-1) = 0

Both 1 and -1 are the zeros of p(x).

4. p(x) = (x + 1)(x – 2)

x = – 1, 2

p(- 1) = (-1 + 1) (- 1 – 2)

= (0) (-3) = 0

p( 2) = (2 + 1) (2 – 2)

= 3 x 0 = 0

As p(- 1) = 0 and p(2) = 0,

Both -1 and 2 are the zeros of p(x).

5. p(x) = x^{2}

x = 0

P(0) = (0)^{2} = 0

As p(0) = 0, 0 is a zero of p(x)

6. p(x) = lx + m, x = – \(\frac{m}{l}\)

p(- \(\frac{m}{l}\)) = l x – \(\frac{m}{l}\) + m

As p(- \(\frac{m}{l}\)) = 0, (-\(\frac{m}{l}\)) is a zero of p(x).

7.

8.

Question 4.

Find the zero of the polynomial in each of the following cases:

- p(x) = x + 5
- p(x) = x – 5
- p(x) = 2x + 5
- p(x) = 3x – 2
- p(x) = 3x
- p(x) = ax, a ≠ 0
- p(x) = cx + d, c ≠ 0, c, d are real numbers.

Solution:

1. p(x) = x + 5

p(x) = 0

x + 5 = 0

x = – 5

∴ – 5 is a zero of the polynomial p(x).

2. P(x) = x – 5

p(x) = 0

x – 5 = 0 x = 5

∴ 5 is a zero of the polynomial p(x).

3. p(x) = 2x + 5

2x + 5 = 0

x = – \(\frac{5}{2}\)

∴ – \(\frac{5}{2}\) is the zero of p(x).

4. p(x) = 3x – 2

3x – 2 = 0

x = \(\frac{2}{3}\)

∴ \(\frac{2}{3}\) is the zero of p(x)

5. p(x) = 3x

3x = 0

x = \(\frac{0}{3}\)

∴ 0 is the zero of p(x).

6. p(x) = ax where a ≠ 0

ax = 0

x = \(\frac{0}{a}\)

∴ 0 is the zero of p(x).

7. p(x) = cx + d where c ≠ 0, c, d are real numbers

cx + d = 0

x = – \(\frac{d}{c}\)

∴ – \(\frac{d}{c}\) is the zero of p(x).

Remainder Theorem:

If a polynomial p{x) is divided by another polynomial (x – a), the remainder isp(a). Where p(x) is any polynomial of degree greater than or equal to one and also degree p(x) > degree p(a) ‘a’ be any real number.

Dividend = divisor x quotient + remainder

Proof:

Let p(x) be any polynomial with degree greater than or equal to 1, suppose that when p(x) is divided by x – a, the quotient is q(x) and remainder is r(x).

p(x) – (x – a) x q(x) + r(x)

p(x) = (x – a) x q(x) + r(x)…..(i)

Case I:

If r(x) = 0

Equation reduce to

p(x) = (x – a) q(x) ……(ii)

On putting x = a, in (ii),

We get p(a) = (a – a) q(a)

p(a) = 0 = Remainder

Case II:

If r(x) ≠ 0

Equation (i) reduced to

p(x) = (x – a) q(x) + r …..(iii)

On putting x = a in (iii), we get

p(a) =(a – a) x q(x) + r

p(a) = r = Remainder

∴ So the remainder is p(a) when p(x) is divided by (x – a).