Exercise 2B RS Aggarwal Class 9 Polynomials Best Solutions

Exercise 2B RS Aggarwal Class 9 contains a total of nine questions. The questions are based on the following topics.

Value of a polynomial
Zeros of a Polynomial

Exercise 2B RS Aggarwal Class 9 Solutions

The first question in Exercise 2B RS Aggarwal Class 9 revolves around finding the value of the given polynomial at a given point.

1. If $p (x) = 5 - 4 x + 2 x^{2}$ , find (i) $p (0)$ , (ii) $p (3)$ , (iii) $p (- 2)$ .

Answer

We have
$p (x) = 5 - 4 x + 2 x^{2}$ .

(i) Putting $x = 0$ , we get.
$p (0) = 5 - 4 \times 0 + 2 \times 0^{2}$ $= 5 - 0 + 0$ $= 5$ .

(ii) Putting $x = 3$ , we get.
$p (3) = 5 - 4 \times 3 + 2 \times 3^{2}$ = $5 - 12 + 18$ = $= 11$ .

(iii) Putting $x = - 2$ , we get.
$p (- 2) = 5 - 4 \times (- 2) + 2 \times {(- 2)}^{2}$ $= 5 + 8 + 8$ $= 21$ .

The second question of Exercise 1F RS Aggarwal Class 9 asks us to find the value of the polynomial at the given points.

2. If $p (y) = 4 + 3 y - y^{2} + 5 y^{3}$ , find (i) $p (0)$ , (ii) $p (2)$ , (iii) $p (- 1)$ .

Answer

We have
$p (y) = 4 + 3 y - y^{2} + 5 y^{3}$ .

(i) Putting $y = 0$ , we get.
$p (0) = 4 + 3 \times 0 - 0^{2} + 5 \times 0^{3}$ $= 4 + 0 - 0$ $= 4$ .

(ii) Putting $y = 2$ , we get.
$p (2) = 4 + 3 \times 2 - 2^{2} + 5 \times 2^{3}$ $= 4 + 6 - 4 + 40$ $= 46$ .

(iii) Putting $y = - 1$ , we get.
$p (- 1) = 4 + 3 \times (- 1) - {(- 1)}^{2} + 5 {(- 1)}^{3}$ $= 4 - 3 - 1 - 5$ $= - 5$ .

The third question of Exercise 2B RS Aggarwal Class 9 asks to find the value of a polynomial at a point.

3. If $f (t) = 4 t^{2} - 3 t + 6$ , find (i) $f (0)$ , (ii) $f (4)$ , (iii) $f (- 5)$ .

Answer

We have
$f (t) = 4 t^{2} - 3 t + 6$ .

(i) Putting $t = 0$ , we get.
$f (0) = 4 \times 0^{2} - 3 \times 0 + 6$ $= 0 - 0 + 6$ $= 6$ .

(ii) Putting $t = 4$ , we get.
$f (4) = 4 \times 4^{2} - 3 \times 4 + 6$ $= 64 - 12 + 6$ $= 58$ .

(iii) Putting $t = - 5$ , we get.
$f (- 5) = 4 \times {(- 5)}^{2} - 3 \times (- 5) + 6$ $= 4 \times 25 + 15 + 6$ $= 121$ .

The fourth question of Exercise 2B RS Aggarwal Class 9 asks us to find the values at different points.

4. If $p (x) = x^{3} - 3 x^{2} + 2 x$ , find $p (0)$ , $p (1)$ , $p (2)$ . What do you conclude?

Answer

We have
$p (x) = x^{3} - 3 x^{2} + 2 x$

(i) Putting $x = 0$ , we get.
$p (0) = 0^{3} - 3 \times 0^{2} + 2 \times 0$ $= 0 - 0 + 0$ $= 0$ .

(ii) Putting $x = 1$ , we get.
$p (1) = 1^{3} - 3 \times 1^{2} + 2 \times 1$ $= 1 - 3 + 2$ $= 0$

(iii) Putting $x = 2$ , we get.
$p (2) = 2^{3} - 3 \times 2^{2} + 2 \times 2$ $= 8 - 12 + 4$ $= 0$

Conclusion:

∵ $p (0) = 0, p (1) = 0, p (2) = 0$
∴ $x = 0, x = 1, x = 2$ are zeros of the polynomial $p (x) = x^{3} - 3 x^{2} + 2 x$ .

The fifth question of Exercise 2B RS Aggarwal Class 9 requires us to determine the value of a function at a specific point and ascertain whether it is equal to zero.

5. If $p (x) = x^{3} + x^{2} - 9 x - 9$ , find $p (0)$ , $p (3)$ , $p (−3)$ and $p (−1)$ . What do you conclude about the zeros of $p (x)$ ? Is 0 a zero of $p (x)$ ?

Answer

We have
$p (x) = x^{3} + x^{2} - 9 x - 9$ .

(i) Putting $x = 0$ , we get.
$p (0) = 0^{3} + 0^{2} - 9 \times 0 - 9$ $= 0 + 0 - 0 - 9$ $= - 9$ .

(ii) Putting $x = 3$ , we get.
$p (3) = 3^{3} + 3^{2} - 9 \times 3 - 9$ $= 27 + 9 - 27 - 9$ $= 0$ .

(iii) Putting $x = −3$ , we get.
$p (- 3) = {(- 3)}^{3} + {(- 3)}^{2} - 9 \times (- 3) - 9$ $= - 27 + 9 + 27 - 9$ $= 0$ .

(iv) Putting $x = −1$ , we get.
$p (- 1) = {(- 1)}^{3} + {(- 1)}^{2} - 9 \times (- 1) - 9$ $= - 1 + 1 + 9 - 9$ $= 0$ .

Conclusion:

$∵ p (3) = 0, p (- 3) = 0, p (- 1) = 0$
$∴ x = 0, x = 1, x = 2$ are the zeros of the polynomial $p (x) = x^{3} + x^{2} - 9 x - 9$ .

$∵ p (0) = - 9 \neq 0,$
$∴ x = 0$ is not a zero of the polynomial $p (x) = x^{3} + x^{2} - 9 x - 9$ .

In Exercise 2B RS Aggarwal Class 9, the sixth question asks us to verify whether a number is a zero of a given polynomial or not.

6. Verify that.

(i) 4 is a zero of the polynomial, $p (x) = x - 4$

Answer

We have
$p (x) = x - 4$

Putting $x = 4$ , we get.
$p (4) = 4 - 4 = 0$ .

$∵ p (4) = 0,$
$∴ x = 4$ is a zero of the polynomial.

Hence, verified.

(ii) −3 is a zero of the polynomial, $q (x) = x + 3$

Answer

We have
$q (x) = x + 3$

Putting $x = - 3$ , we get.
$q (- 3) = - 3 + 3 = 0$ .

$∵ q (- 3) = 0,$
$∴ x = - 3$ is a zero of the polynomial.

Hence, verified.

(iii) $\frac{2}{5}$ is a zero of the polynomial, $f (x) = 2 - 5 x$ .

Answer

We have
$f (x) = 2 - 5 x$ .

Putting $x = \frac{2}{5}$ , we get.
$f (\frac{2}{5}) =$ $2 - 5 \times \frac{2}{5}$ $= 2 - 2 = 0$ .

$∵ f (\frac{2}{5}) = 0,$
$∴ x = \frac{2}{5}$ is a zero of the polynomial.

Hence, verified.

(iv) $\frac{−1}{2}$ is a zero of the polynomial, $g (y) = 2 y + 1$ .

Answer

We have
$g (y) = 2 y + 1$ .

Putting $y = \frac{- 1}{2}$ , we get.
$g (\frac{- 1}{2}) =$ $2 \times (\frac{- 1}{2}) + 1$ $= - 1 + 1$ $= 0$

$∵ g (\frac{- 1}{2}) = 0,$
$∴ y = \frac{- 1}{2}$ is a zero of the polynomial.

Hence, verified.

In Exercise 2B RS Aggarwal Class 9, the seventh question asks us to verify if the given values are zeros of the given polynomial.

7. Verify that.

(i) 1 and 2 are the zeros of the polynomial, $p (x) = x^{2} - 3 x + 2$ .

Answer

The given polynomial is
$p (x) = x^{2} - 3 x + 2$ .

Putting $x = 1$ , we get.
$p (1) = 1^{2} - 3 \times 1 + 2$ $= 1 - 3 + 2$ $= 0$

Putting $x = 2$ , we get.
$p (2) = 2^{2} - 3 \times 2 + 2$ $= 4 - 6 + 2$ $= 0$

$∵ p (1) = 0, p (2) = 0$
$∴ x = 1 a n d x = 2 a r e z e r o e s o f p (x) .$

Hence, verified.

(ii) 2 and −3 are the zeros of the polynomial, $q (x) = x^{2} + x - 6$ .

Answer

The given polynomial is
$q (x) = x^{2} + x - 6$ .

Putting $x = 2$ , we get.
$q (2) = 2^{2} + 2 - 6$ $= 4 + 2 - 6$ $= 0$

Putting $x = - 3$ , we get.
$q (- 3) = {(- 3)}^{2} + (- 3) - 6$ $= 9 - 3 - 6$ $= 0$

$∵ q (2) = 0, q (- 3) = 0$
$∴ x = 2 a n d x = - 3 a r e z e r o e s o f q (x) .$

Hence, verified.

(iii) 0 and 3 are the zeros of the polynomial, $r (x) = x^{2} - 3 x$ .

Answer

The given polynomial is
$r (x) = x^{2} - 3 x$ .

Putting $x = 0$ , we get.
$r (0) = 0^{2} - 3 \times 0$ $= 0 - 0$ $= 0$

Putting $x = 3$ , we get.
$r (3) = 3^{2} - 3 \times 3$ $= 9 - 9$ $= 0$

$∵ r (0) = 0, r (3) = 0$
$∴ x = 0 a n d x = 3 a r e z e r o e s o f r (x) .$

Hence, verified.

In Exercise 2B RS Aggarwal Class 9, the eighth question asks us to find the zeros of the polynomials.

8. Find the zero of the polynomial:

(i) $p (x) = x - 5$

Answer

The given polynomial is
$p (x) = x - 5$ .

Let $a$ be the zero of $p (x)$ .
∴ $p (a) = 0$ .
$\Rightarrow a - 5 = 0$ .
$\Rightarrow a = 5$ .

Hence, the zero of the polynomial is 5.

(ii) $q (x) = x + 4$

Answer

The given polynomial is
$q (x) = x + 4$

Let $a$ be the zero of $q (x)$ .
∴ $q (a) = 0$ .
$\Rightarrow 4 a = 0$
$\Rightarrow a = 0$

Hence, the zero of the polynomial is $- 4$ .

(iii) $r (x) = 2 x + 5$

Answer

The given polynomial is
$r (x) = 2 x + 5$

Let $a$ be the zero of $r (x)$ .
∴ $r (a) = 0$ .
$\Rightarrow 2 a + 5 = 0$ .
$\Rightarrow a = \frac{- 5}{2}$

Hence, the zero of the polynomial is $\frac{- 5}{2}$ .

(iv) $f (x) = 3 x + 1$

Answer

The given polynomial is
$f (x) = 3 x + 1$ .

Let $a$ be the zero of $f (x)$ .
∴ $f (a) = 0$ .
$\Rightarrow 3 a + 1 = 0$ .
$\Rightarrow a = \frac{- 1}{3}$

Hence, the zero of the polynomial is $\frac{- 1}{3}$ .

(v) $g (x) = 5 - 4 x$

Answer

The given polynomial is
$g (x) = 5 - 4 x$

Let $a$ be the zero of $g (x)$ .
∴ $g (a) = 0$ .
$\Rightarrow 5 - 4 a = 0$
$\Rightarrow a = \frac{5}{4}$

Hence, the zero of the polynomial is $\frac{5}{4}$ .

(vi) $h (x) = 6 x - 2$

Answer

The given polynomial is
$h (x) = 6 x - 2$ .

Let $a$ be the zero of $h (x)$ .
∴ $h (a) = 0$ .
$\Rightarrow 6 a - 2 = 0$
$\Rightarrow a = \frac{2}{6} = \frac{1}{3}$ .

Hence, the zero of the polynomial is $\frac{1}{3}$ .

(vii) $p (x) = a x, a \neq 0$

Answer

The given polynomial is
$p (x) = a x, a \neq 0$

Let $z$ be the zero of $p (x)$ .
∴ $p (z) = 0$ .
$\Rightarrow a z = 0$ .
$\Rightarrow z = 0$ .

Hence, the zero of the polynomial is 0.

(viii) $q (x) = 4 x$

Answer

The given polynomial is
$q (x) = 4 x$ .

Let $a$ be the zero of $q (x)$ .
∴ $q (a) = 0$ .
$\Rightarrow 4 a = 0$ .
$\Rightarrow a = 0$ .

Hence, the zero of the polynomial is 0.

In Exercise 2B RS Aggarwal Class 9, the ninth question asks us to find the value of $a$ and $b$ from the given polynomial.

9. If 2 and 0 are the zeros of the polynomial $f (x) = 2 x^{3} - 5 x^{2} + a x + b$ then find the value of $a$ and $b$ .

Answer

The given polynomial is
$f (x) = 2 x^{3} - 5 x^{2} + a x + b$ .

$∵ x = 2 a n d x = 0 a r e z e r o s o f f (x)$

$∴ f (2) = 0$
$\Rightarrow 2 \times 2^{3} - 5 \times 2^{2} + 2 a + b = 0$
$\Rightarrow 16 - 20 + 2 a + b = 0$
$\Rightarrow - 4 + 2 a + b = 0$
$\Rightarrow 2 a + b = 4 . . . (i)$

Also, $f (0) = 0$
$\Rightarrow 2 \times 0^{3} - 5 \times 0^{2} + a \times 0 + b = 0$
$\Rightarrow 0 - 0 + 0 + b = 0$
$\Rightarrow b = 0$

Putting $b = 0$ in equation $(i)$ , we get.
$\Rightarrow 2 a + 0 = 4$
$\Rightarrow 2 a = 4$
$\Rightarrow a = 2$ .

Hence, $a = 2$ and $b = 0$ .

Exercise 2B RS Aggarwal Class 9 Solutions

1. If p(x)=5−4x+2x2, find (i) p(0), (ii) p(3), (iii) p(−2).

Answer

2. If p(y)=4+3y−y2+5y3, find (i) p(0), (ii) p(2), (iii) p(−1).

Answer

3. If f(t)=4t2−3t+6, find (i) f(0), (ii) f(4), (iii) f(−5).

Answer

4. If p(x)=x3−3x2+2x, find p(0), p(1), p(2). What do you conclude?

Answer

5. If p(x)=x3+x2−9x−9, find p(0), p(3), p(−3) and p(−1). What do you conclude about the zeros of p(x)? Is 0 a zero of p(x)?

Answer

6. Verify that.

(i) 4 is a zero of the polynomial, p(x)=x−4

Answer

(ii) −3 is a zero of the polynomial, q(x)=x+3

Answer

(iii) 25 is a zero of the polynomial, f(x)=2−5x.

Answer

(iv) −12 is a zero of the polynomial, g(y)=2y+1.

Answer

7. Verify that.

(i) 1 and 2 are the zeros of the polynomial, p(x)=x2-3x+2.

Answer

(ii) 2 and −3 are the zeros of the polynomial, q(x)=x2+x−6.

Answer

(iii) 0 and 3 are the zeros of the polynomial, r(x)=x2−3x.

Answer

8. Find the zero of the polynomial:

(i) p(x)=x−5

Answer

(ii) q(x)=x+4

Answer

(iii) r(x)=2x+5

Answer

(iv) f(x)=3x+1

Answer

(v) g(x)=5−4x

Answer

(vi) h(x)=6x−2

Answer

(vii) p(x)=ax,a≠0

Answer

(viii) q(x)=4x

Answer

9. If 2 and 0 are the zeros of the polynomial f(x)=2x3−5x2+ax+b then find the value of a and b.

Answer

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1. If $p (x) = 5 - 4 x + 2 x^{2}$ , find (i) $p (0)$ , (ii) $p (3)$ , (iii) $p (- 2)$ .

2. If $p (y) = 4 + 3 y - y^{2} + 5 y^{3}$ , find (i) $p (0)$ , (ii) $p (2)$ , (iii) $p (- 1)$ .

3. If $f (t) = 4 t^{2} - 3 t + 6$ , find (i) $f (0)$ , (ii) $f (4)$ , (iii) $f (- 5)$ .

4. If $p (x) = x^{3} - 3 x^{2} + 2 x$ , find $p (0)$ , $p (1)$ , $p (2)$ . What do you conclude?

5. If $p (x) = x^{3} + x^{2} - 9 x - 9$ , find $p (0)$ , $p (3)$ , $p (−3)$ and $p (−1)$ . What do you conclude about the zeros of $p (x)$ ? Is 0 a zero of $p (x)$ ?

(i) 4 is a zero of the polynomial, $p (x) = x - 4$

(ii) −3 is a zero of the polynomial, $q (x) = x + 3$

(iii) $\frac{2}{5}$ is a zero of the polynomial, $f (x) = 2 - 5 x$ .

(iv) $\frac{−1}{2}$ is a zero of the polynomial, $g (y) = 2 y + 1$ .

(i) 1 and 2 are the zeros of the polynomial, $p (x) = x^{2} - 3 x + 2$ .

(ii) 2 and −3 are the zeros of the polynomial, $q (x) = x^{2} + x - 6$ .

(iii) 0 and 3 are the zeros of the polynomial, $r (x) = x^{2} - 3 x$ .

(i) $p (x) = x - 5$

(ii) $q (x) = x + 4$

(iii) $r (x) = 2 x + 5$

(iv) $f (x) = 3 x + 1$

(v) $g (x) = 5 - 4 x$

(vi) $h (x) = 6 x - 2$

(vii) $p (x) = a x, a \neq 0$

(viii) $q (x) = 4 x$

9. If 2 and 0 are the zeros of the polynomial $f (x) = 2 x^{3} - 5 x^{2} + a x + b$ then find the value of $a$ and $b$ .