RS AGGARWAL CLASS 9 Chapter 2 POLYNOMIALS EXERCISE 2B

 EXERCISE 2B

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Question 1:

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

Answer 1:

$\left(i\right)p\left(x\right)=5-4x+2x^2\Rightarrow p\left(0\right)=\left(5-4\times 0+2\times 0^2\right)$$$\begin{gathered} \left(\mathrm{i}\right) p\left(x\right)=5-4x+2x^2 \\ \Rightarrow p\left(0\right)=\left(5-4\times 0+2\times 0^2\right) \end{gathered}$$
        $=\left(5-0+0\right)=5$$$\begin{gathered} =\left(5-0+0\right) \\ =5 \end{gathered}$$

$\left(\mathrm{ii}\right)p\left(x\right)=5-4x+2x^2\Rightarrow p\left(3\right)=\left(5-4\times 3+2\times 3^2\right)$$$\begin{gathered} \left(\mathrm{ii}\right) p\left(x\right)=5-4x+2x^2 \\ \Rightarrow p\left(3\right)=\left(5-4\times 3+2\times 3^2\right) \end{gathered}$$
        $=\left(5-12+18\right)=11$$$\begin{gathered} =\left(5-12+18\right) \\ =11 \end{gathered}$$

$\left(\mathrm{iii}\right)p\left(x\right)=5-4x+2x^2\Rightarrow p\left(-2\right)=\left[5-4\times \left(-2\right)+2\times {\left(-2\right)}^2\right]$$$\begin{gathered} \left(\mathrm{iii}\right) p\left(x\right)=5-4x+2x^2 \\ \Rightarrow p\left(-2\right)=\left[5-4\times \left(-2\right)+2\times {\left(-2\right)}^2\right] \end{gathered}$$           
          $=\left(5+8+8\right)=21$$$\begin{gathered} =\left(5+8+8\right) \\ =21 \end{gathered}$$

Question 2:

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

Answer 2:

$\left(i\right)p\left(y\right)=4+3y-y^2+5y^3\Rightarrow p\left(0\right)=\left(4+3\times 0-0^2+5\times 0^3\right)$$$\begin{gathered} \left(\mathrm{i}\right) p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(0\right)=\left(4+3\times 0-0^2+5\times 0^3\right) \end{gathered}$$
        $=\left(4+0-0+0\right)=4$$$\begin{gathered} =\left(4+0-0+0\right) \\ =4 \end{gathered}$$

$\left(\mathrm{ii}\right)p\left(y\right)=4+3y-y^2+5y^3\Rightarrow p\left(2\right)=\left(4+3\times 2-2^2+5\times 2^3\right)$$$\begin{gathered} \left(\mathrm{ii}\right) p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(2\right)=\left(4+3\times 2-2^2+5\times 2^3\right) \end{gathered}$$
        $=\left(4+6-4+40\right)=46$$$\begin{gathered} =\left(4+6-4+40\right) \\ =46 \end{gathered}$$

$\left(\mathrm{iii}\right)p\left(y\right)=4+3y-y^2+5y^3\Rightarrow p\left(-1\right)=\left[4+3\times \left(-1\right)-{\left(-1\right)}^2+5\times {\left(-1\right)}^3\right]$$$\begin{gathered} \left(\mathrm{iii}\right) p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(-1\right)=\left[4+3\times \left(-1\right)-{\left(-1\right)}^2+5\times {\left(-1\right)}^3\right] \end{gathered}$$
          $=\left(4-3-1-5\right)=-5$$$\begin{gathered} =\left(4-3-1-5\right) \\ =-5 \end{gathered}$$

Question 3:

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

Answer 3:

$\left(i\right)f\left(t\right)=4t^2-3t+6\Rightarrow f\left(0\right)=\left(4\times 0^2-3\times 0+6\right)$$$\begin{gathered} \left(\mathrm{i}\right) f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(0\right)=\left(4\times 0^2-3\times 0+6\right) \end{gathered}$$
       $=\left(0-0+6\right)=6$$$\begin{gathered} =\left(0-0+6\right) \\ =6 \end{gathered}$$

$\left(\mathrm{ii}\right)f\left(t\right)=4t^2-3t+6\Rightarrow f\left(4\right)=\left(4\times 4^2-3\times 4+6\right)$$$\begin{gathered} \left(\mathrm{ii}\right) f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(4\right)=\left(4\times 4^2-3\times 4+6\right) \end{gathered}$$
        $=\left(64-12+6\right)=58$$$\begin{gathered} =\left(64-12+6\right) \\ =58 \end{gathered}$$

$\left(\mathrm{iii}\right)f\left(t\right)=4t^2-3t+6\Rightarrow f\left(-5\right)=\left[4\times {\left(-5\right)}^2-3\times \left(-5\right)+6\right]$$$\begin{gathered} \left(\mathrm{iii}\right) f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(-5\right)=\left[4\times {\left(-5\right)}^2-3\times \left(-5\right)+6\right] \end{gathered}$$
          $=\left(100+15+6\right)=121$$$\begin{gathered} =\left(100+15+6\right) \\ =121 \end{gathered}$$

Question 4:

If $p\left(x\right)=x^3-3x^2+2x$$p\left(x\right)=x^3-3x^2+2x$, find p(0), p(1), p(2). What do you conclude?

Answer 4:

$p\left(x\right)=x^3-3x^2+2x$$p\left(x\right)=x^3-3x^2+2x$      .....(1)

Putting x = 0 in (1), we get

$p\left(0\right)=0^3-3\times 0^2+2\times 0=0$$p\left(0\right)=0^3-3\times 0^2+2\times 0=0$

Thus, x = 0 is a zero of p(x).

Putting x = 1 in (1), we get

$p\left(1\right)=1^3-3\times 1^2+2\times 1=1-3+2=0$$p\left(1\right)=1^3-3\times 1^2+2\times 1=1-3+2=0$

Thus, x = 1 is a zero of p(x).

Putting x = 2 in (1), we get

$p\left(2\right)=2^3-3\times 2^2+2\times 2=8-3\times 4+4=8-12+4=0$$p\left(2\right)=2^3-3\times 2^2+2\times 2=8-3\times 4+4=8-12+4=0$

Thus, x = 2 is a zero of p(x).

Question 5:

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

Answer 5:

p(x) = x³ + x² – 9x – 9         .....(1)

Putting x = 0 in (1), we get

p(0) = 0³ + 0² – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0

Thus, x = 0 is not a zero of p(x).

Putting x = 3 in (1), we get

p(3) = 3³ + 3² – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).

Putting x = –3 in (1), we get

p(–3) = (–3)³ + (–3)² – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0

Thus, x = –3 is a zero of p(x).

Putting x = –1 in (1), we get

p(–1) = (–1)³ + (–1)² – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0

Thus, x = –1 is a zero of p(x).

Question 6:

Verify that:
(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}$$\frac{2}{5}$is a zero of the polynomial, f(x) = 2 − 5x.
(iv) $\frac{-1}{2}$$\frac{-1}{2}$is a zero of the polynomial g(y) = 2y + 1.

Answer 6:

$\left(i\right)p\left(x\right)=x-4\Rightarrow p\left(4\right)=4-4$$$\begin{gathered} \left(\mathrm{i}\right) p\left(x\right)=x-4 \\ \Rightarrow p\left(4\right)=4-4 \end{gathered}$$
         = 0
Hence, 4 is the zero of the given polynomial.

$\left(\mathrm{ii}\right)p\left(x\right)=\left(-3\right)+3\Rightarrow p\left(3\right)=0$$$\begin{gathered} \left(\mathrm{ii}\right) p\left(x\right)=\left(-3\right)+3 \\ \Rightarrow p\left(3\right)=0 \end{gathered}$$
Hence, 3 is the zero of the given polynomial.

$\left(\mathrm{iii}\right)p\left(x\right)=2-5x\Rightarrow p\left(\frac{2}{5}\right)=2-5\times \left(\frac{2}{5}\right)$$$\begin{gathered} \left(\mathrm{iii}\right) p\left(x\right)=2-5x \\ \Rightarrow p\left(\frac{2}{5}\right)=2-5\times \left(\frac{2}{5}\right) \end{gathered}$$
          $=2-2=0$$$\begin{gathered} =2-2 \\ =0 \end{gathered}$$

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

$\left(\mathrm{iv}\right)p\left(y\right)=2y+1\Rightarrow p\left(-\frac{1}{2}\right)=2\times \left(-\frac{1}{2}\right)+1=-1+1=0$$$\begin{gathered} \left(\mathrm{iv}\right) p\left(y\right)=2y+1 \\ \Rightarrow p\left(-\frac{1}{2}\right)=2\times \left(-\frac{1}{2}\right)+1 \\ =-1+1 \\ =0 \end{gathered}$$

Hence, $-\frac{1}{2}$$-\frac{1}{2}$ is the zero of the given polynomial.

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Question 7:

Verify that
(i) 1 and 2 are the zeros of the polynomial p(x) = x² − 3x + 2.
(ii) 2 and −3 are the zeros of the polynomial q(x) = x² + x − 6.
(iii) 0 and 3 are the zeros of the polynomial r(x) = x² − 3x.

Answer 7:

$\left(i\right)p\left(x\right)=x^2-3x+2=\left(x-1\right)\left(x-2\right)\Rightarrow p\left(1\right)=\left(1-1\right)\times \left(1-2\right)$$$\begin{gathered} \left(\mathrm{i}\right) p\left(x\right)=x^2-3x+2=\left(x-1\right)\left(x-2\right) \\ \Rightarrow p\left(1\right)=\left(1-1\right)\times \left(1-2\right) \end{gathered}$$
        $=0\times \left(-1\right)=0$$$\begin{gathered} =0\times \left(-1\right) \\ =0 \end{gathered}$$
Also,
$p\left(2\right)=\left(2-1\right)\left(2-2\right)$$p\left(2\right)=\left(2-1\right)\left(2-2\right)$
     $=\left(-1\right)\times 0=0$$$\begin{gathered} =\left(-1\right)\times 0 \\ =0 \end{gathered}$$

Hence, 1 and 2 are the zeroes of the given polynomial.

$\left(\mathrm{ii}\right)p\left(x\right)=x^2+x-6\Rightarrow p\left(2\right)=2^2+2-6$$$\begin{gathered} \left(\mathrm{ii}\right) p\left(x\right)=x^2+x-6 \\ \Rightarrow p\left(2\right)=2^2+2-6 \end{gathered}$$
        $=4-4=0$$$\begin{gathered} =4-4 \\ =0 \end{gathered}$$
Also,
$p\left(-3\right)={\left(-3\right)}^2+\left(-3\right)-6=9-9=0$$$\begin{gathered} p\left(-3\right)={\left(-3\right)}^2+\left(-3\right)-6 \\ =9-9 \\ =0 \end{gathered}$$

Hence, 2 and $-3$$-3$ are the zeroes of the given polynomial.

$\left(\mathrm{iii}\right)p\left(x\right)=x^2-3x\Rightarrow p\left(0\right)=0^2-3\times 0$$$\begin{gathered} \left(\mathrm{iii}\right) p\left(x\right)=x^2-3x \\ \Rightarrow p\left(0\right)=0^2-3\times 0 \end{gathered}$$

Also,
$p\left(3\right)=3^2-3\times 3=9-9=0$$$\begin{gathered} p\left(3\right)=3^2-3\times 3 \\ =9-9 \\ =0 \end{gathered}$$

Hence, 0 and 3 are the zeroes of the given polynomial.

Question 8:

Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) r(x) = 2x + 5
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 2
(vii) p(x) = ax, a ≠ 0
(viii) q(x) = 4x

Answer 8:

$\left(i\right)p\left(x\right)=0\Rightarrow x-5=0\Rightarrow x=5\mathrm{Hence},5\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(x\right).\left(\mathrm{ii}\right)q\left(x\right)=0\Rightarrow x+4=0\Rightarrow x=-4\mathrm{Hence},-4\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}q\left(x\right).\left(\mathrm{iii}\right)r\left(x\right)=0\Rightarrow 2x+5=0\Rightarrow t=\frac{-5}{2}\mathrm{Hence},\frac{-5}{2}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(t\right).$$$\begin{gathered} \left(\mathrm{i}\right) p\left(x\right)=0\Rightarrow x-5=0 \\ \Rightarrow x=5 \\ \mathrm{Hence}, 5 \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} p\left(x\right). \\ \left(\mathrm{ii}\right) q\left(x\right)=0\Rightarrow x+4=0 \\ \Rightarrow x=-4 \\ \mathrm{Hence}, -4 \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} q\left(x\right). \\ \left(\mathrm{iii}\right) r\left(x\right)=0\Rightarrow 2x+5=0 \\ \Rightarrow t=\frac{-5}{2} \\ \mathrm{Hence}, \frac{-5}{2} \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} p\left(t\right). \end{gathered}$$

$\left(\mathrm{iv}\right)f\left(x\right)=0\Rightarrow 3x+1=0\Rightarrow x=-\frac{1}{3}\mathrm{Hence},-\frac{1}{3}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}f\left(x\right).\left(v\right)g\left(x\right)=0\Rightarrow 5-4x=0\Rightarrow x=\frac{5}{4}\mathrm{Hence},\frac{5}{4}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}g\left(x\right).\left(\mathrm{vi}\right)h\left(x\right)=0\Rightarrow 6x-2=0\Rightarrow x=\frac{2}{6}=\frac{1}{3}\mathrm{Hence},\frac{1}{3}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}h\left(x\right).$$$\begin{gathered} \left(\mathrm{iv}\right) f\left(x\right)=0\Rightarrow 3x+1=0 \\ \Rightarrow x=-\frac{1}{3} \\ \mathrm{Hence}, -\frac{1}{3} \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} f\left(x\right). \\ \left(\mathrm{v}\right) g\left(x\right)=0\Rightarrow 5-4x=0 \\ \Rightarrow x=\frac{5}{4} \\ \mathrm{Hence}, \frac{5}{4} \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} g\left(x\right). \\ \left(\mathrm{vi}\right) h\left(x\right)=0\Rightarrow 6x-2=0 \\ \Rightarrow x=\frac{2}{6}=\frac{1}{3} \\ \mathrm{Hence}, \frac{1}{3} \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} h\left(x\right). \end{gathered}$$

$\left(\mathrm{vii}\right)p\left(x\right)=0\Rightarrow ax=0\Rightarrow x=0\mathrm{Hence},0\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(x\right).\left(\mathrm{viii}\right)q\left(x\right)=0\Rightarrow 4x=0\Rightarrow x=0\mathrm{Hence},0\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}q\left(x\right).$$$\begin{gathered} \left(\mathrm{vii}\right) p\left(x\right)=0\Rightarrow ax=0 \\ \Rightarrow x=0 \\ \mathrm{Hence}, 0 \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} p\left(x\right). \\ \left(\mathrm{viii}\right) q\left(x\right)=0\Rightarrow 4x=0 \\ \Rightarrow x=0 \\ \mathrm{Hence}, 0 \mathrm{is} \mathrm{the} \mathrm{zero} \mathrm{of} \mathrm{the} \mathrm{polynomial} q\left(x\right). \end{gathered}$$

Question 9:

If 2 and 0 are the zeros of the polynomial $f\left(x\right)=2x^3-5x^2+ax+b$$f\left(x\right)=2x^3-5x^2+ax+b$ then find the values of a and b.
Hint f(2) = 0 and f(0) = 0.

Answer 9:

It is given that 2 and 0 are the zeroes of the polynomial $f\left(x\right)=2x^3-5x^2+ax+b$$f\left(x\right)=2x^3-5x^2+ax+b$.
∴ f(2) = 0
$\Rightarrow 2\times 2^3-5\times 2^2+a\times 2+b=0\Rightarrow 16-20+2a+b=0\Rightarrow -4+2a+b=0\Rightarrow 2a+b=4.....\left(1\right)$$$\begin{gathered} \Rightarrow 2\times 2^3-5\times 2^2+a\times 2+b=0 \\ \Rightarrow 16-20+2a+b=0 \\ \Rightarrow -4+2a+b=0 \\ \Rightarrow 2a+b=4 .....\left(1\right) \end{gathered}$$
Also,
f(0) = 0
$\Rightarrow 2\times 0^3-5\times 0^2+a\times 0+b=0\Rightarrow 0-0+0+b=0\Rightarrow b=0$$$\begin{gathered} \Rightarrow 2\times 0^3-5\times 0^2+a\times 0+b=0 \\ \Rightarrow 0-0+0+b=0 \\ \Rightarrow b=0 \end{gathered}$$
Putting b = 0 in (1), we get
$2a+0=4\Rightarrow 2a=4\Rightarrow a=2$$$\begin{gathered} 2a+0=4 \\ \Rightarrow 2a=4 \\ \Rightarrow a=2 \end{gathered}$$
Thus, the values of a and b are 2 and 0, respectively.