GSEB Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.2

Question 1. Find the value of the polynomial \( 5x - 4x^2 + 3 \) at
(i) \( x = 0 \)
(ii) \( x = -1 \)
(iii) \( x = 2 \)
Answer:
Let \( p(x) = 5x - 4x^2 + 3 \).
(i) Let's substitute \( x = 0 \) into \( p(x) \):
\( p(0) = 5 \times (0) - 4 \times (0)^2 + 3 \)
\( p(0) = 3 \).
(ii) Next, we substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = 5 \times (-1) - 4(-1)^2 + 3 \)
\( = -5 - 4 + 3 \)
\( = -9 + 3 \)
\( = -6 \).
(iii) Finally, we substitute \( x = 2 \) into \( p(x) \):
\( p(2) = 5 \times 2 - 4(2)^2 + 3 \)
\( = 10 - 4 \times 4 + 3 \)
\( = 10 - 16 + 3 \)
\( = 13 - 16 \)
\( = -3 \).
In simple words: To find the value, we replace the variable \( x \) with the given numbers in the polynomial expression and then calculate the result. This helps us see what number the polynomial becomes for each specific \( x \) value.

Question 2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) \( p(y) = y^2 - y + 1 \)
(ii) \( p(t) = 2 + t + 2t^2 - t^3 \)
(iii) \( p(x) = x^3 \)
(iv) \( p(x) = (x - 1) (x + 1) \)
Answer:
(i) For \( p(y) = y^2 - y + 1 \):
When \( y = 0 \): \( p(0) = (0)^2 - (0) + 1 = 1 \).
When \( y = 1 \): \( p(1) = (1)^2 - (1) + 1 = 1 - 1 + 1 = 1 \).
When \( y = 2 \): \( p(2) = (2)^2 - (2) + 1 = 4 - 2 + 1 = 3 \).
(ii) For \( p(t) = 2 + t + 2t^2 - t^3 \):
When \( t = 0 \): \( p(0) = 2 + (0) + 2(0)^2 - (0)^3 = 2 \).
When \( t = 1 \): \( p(1) = 2 + (1) + 2(1)^2 - (1)^3 = 2 + 1 + 2 - 1 = 4 \).
When \( t = 2 \): \( p(2) = 2 + 2 + 2(2)^2 - (2)^3 = 4 + 2 \times 4 - 8 = 4 + 8 - 8 = 4 \).
(iii) For \( p(x) = x^3 \):
When \( x = 0 \): \( p(0) = (0)^3 = 0 \).
When \( x = 1 \): \( p(1) = (1)^3 = 1 \).
When \( x = 2 \): \( p(2) = (2)^3 = 8 \).
(iv) For \( p(x) = (x - 1) (x + 1) \):
When \( x = 0 \): \( p(0) = (0 - 1) (0 + 1) = (-1) \times (1) = -1 \).
When \( x = 1 \): \( p(1) = (1 - 1) (1 + 1) = 0 \times 2 = 0 \).
When \( x = 2 \): \( p(2) = (2 - 1) (2 + 1) = 1 \times 3 = 3 \).
In simple words: To find \( p(0), p(1), \) and \( p(2) \), simply replace the variable in each polynomial (like \( y, t, \) or \( x \)) with 0, 1, and 2, and then calculate the resulting value. This shows the polynomial's output for those specific input numbers.

Question 3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) \( p(x) = 3x + 1, x = \frac { 1 }{ 3 } \)
(ii) \( p(x) = 5x - \pi, x = \frac { 4 }{ 5 } \)
(iii) \( p(x) = x^2 - 1, x = 1, -1 \)
(iv) \( p(x) = (x + 1) (x - 2), x = -1, 2 \)
(v) \( p(x) = x^2, x = 0 \)
(vi) \( p(x) = lx+m, x = \frac { -m }{l} \)
(vii) \( p(x) = 3x^2 - 1, x = \frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)
(viii) \( p(x) = 2x + 1, x = \frac { 1 }{ 2 } \)
Answer:
(i) For \( p(x) = 3x + 1 \), we test \( x = -\frac{1}{3} \) (as per the provided solution steps):
Substitute \( x = -\frac{1}{3} \):
\( p(-\frac{1}{3}) = 3 \times (-\frac{1}{3}) + 1 \)
\( = -1 + 1 \)
\( = 0 \).
Since \( p(-\frac{1}{3}) = 0 \), it implies that \( x = -\frac{1}{3} \) is a zero of \( p(x) \).
(ii) For \( p(x) = 5x - \pi \), we test \( x = \frac{4}{5} \):
Substitute \( x = \frac{4}{5} \):
\( p(\frac{4}{5}) = 5 \times (\frac{4}{5}) - \pi \)
\( = 4 - \pi \).
Since \( 4 - \pi \neq 0 \), it means that \( x = \frac{4}{5} \) is not a zero of \( p(x) \).
(iii) For \( p(x) = x^2 - 1 \), we test \( x = 1 \) and \( x = -1 \):
Substitute \( x = 1 \):
\( p(1) = (1)^2 - 1 = 1 - 1 = 0 \).
Substitute \( x = -1 \):
\( p(-1) = (-1)^2 - 1 = 1 - 1 = 0 \).
Because both \( p(1) = 0 \) and \( p(-1) = 0 \), we know that \( x = 1 \) and \( x = -1 \) are zeroes of \( p(x) \).
(iv) For \( p(x) = (x + 1) (x - 2) \), we test \( x = -1 \) and \( x = 2 \):
Substitute \( x = -1 \):
\( p(-1) = (-1 + 1) (-1 - 2) \)
\( = 0 \times (-3) \)
\( = 0 \).
Substitute \( x = 2 \):
\( p(2) = (2 + 1) (2 - 2) \)
\( = 3 \times (0) \)
\( = 0 \).
Because both \( p(-1) = 0 \) and \( p(2) = 0 \), it shows that \( x = -1 \) and \( x = 2 \) are zeroes of \( p(x) \).
(v) For \( p(x) = x^2 \), we test \( x = 0 \):
Substitute \( x = 0 \):
\( p(0) = (0)^2 = 0 \).
Since \( p(0) = 0 \), it means that \( x = 0 \) is a zero of \( p(x) \).
(vi) For \( p(x) = lx + m \), we test \( x = -\frac{m}{l} \):
Substitute \( x = -\frac{m}{l} \):
\( p(-\frac{m}{l}) = l(-\frac{m}{l}) + m \)
\( = -m + m \)
\( = 0 \).
Since \( p(-\frac{m}{l}) = 0 \), it confirms that \( x = -\frac{m}{l} \) is a zero of \( p(x) \).
(vii) For \( p(x) = 3x^2 - 1 \), we test \( x = \frac{-1}{\sqrt{3}} \) and \( x = \frac{2}{\sqrt{3}} \):
Substitute \( x = \frac{-1}{\sqrt{3}} \):
\( p(\frac{-1}{\sqrt{3}}) = 3(\frac{-1}{\sqrt{3}})^2 - 1 \)
\( = 3(\frac{1}{3}) - 1 \)
\( = 1 - 1 \)
\( = 0 \).
Substitute \( x = \frac{2}{\sqrt{3}} \):
\( p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 - 1 \)
\( = 3(\frac{4}{3}) - 1 \)
\( = 4 - 1 \)
\( = 3 \).
Since \( p(\frac{-1}{\sqrt{3}}) = 0 \), \( x = \frac{-1}{\sqrt{3}} \) is a zero of \( p(x) \). However, since \( p(\frac{2}{\sqrt{3}}) = 3 \neq 0 \), \( x = \frac{2}{\sqrt{3}} \) is not a zero of \( p(x) \).
(viii) For \( p(x) = 2x + 1 \), we test \( x = \frac{1}{2} \):
Substitute \( x = \frac{1}{2} \):
\( p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 \)
\( = 1 + 1 \)
\( = 2 \).
Since \( 2 \neq 0 \), it means that \( x = \frac{1}{2} \) is not a zero of \( p(x) \).
In simple words: To check if a number is a "zero" of a polynomial, you put that number into the polynomial. If the answer you get is zero, then yes, it's a zero. If the answer is anything else, then it's not a zero.

Question 4. Find the zero of the polynomial in each of the following cases:
(i) \( p(x) = x + 5 \)
(ii) \( p(x) = x - 5 \)
(iii) \( p(x) = 2x + 5 \)
(iv) \( p(x) = 3x - 2 \)
(v) \( p(x) = 3x \)
(vi) \( p(x) = ax, a \neq 0 \)
(vii) \( p(x) = cx + d, c \neq 0, c, d \) are real numbers.
Answer:
(i) For \( p(x) = x + 5 \):
Set \( p(x) = 0 \)
\( x + 5 = 0 \)

\( \implies x = -5 \).
Thus, \( x = -5 \) is the zero of polynomial \( p(x) \).
(ii) For \( p(x) = x - 5 \):
Set \( p(x) = 0 \)
\( x - 5 = 0 \)

\( \implies x = 5 \).
Thus, \( x = 5 \) is the zero of polynomial \( p(x) \).
(iii) For \( p(x) = 2x + 5 \):
Set \( p(x) = 0 \)
\( 2x + 5 = 0 \)

\( \implies 2x = -5 \)

\( \implies x = -\frac{5}{2} \).
Thus, \( x = -\frac{5}{2} \) is the zero of polynomial \( p(x) \).
(iv) For \( p(x) = 3x - 2 \):
Set \( p(x) = 0 \)
\( 3x - 2 = 0 \)

\( \implies 3x = 2 \)

\( \implies x = \frac{2}{3} \).
Thus, \( x = \frac{2}{3} \) is the zero of polynomial \( p(x) \).
(v) For \( p(x) = 3x \):
Set \( p(x) = 0 \)
\( 3x = 0 \)

\( \implies x = 0 \).
Thus, \( x = 0 \) is the zero of polynomial \( p(x) \).
(vi) For \( p(x) = ax \), where \( a \neq 0 \):
Set \( p(x) = 0 \)
\( ax = 0 \)

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

\( \implies x = 0 \).
Thus, \( x = 0 \) is the zero of polynomial \( p(x) \).
(vii) For \( p(x) = cx + d \), where \( c \neq 0 \) and \( c, d \) are real numbers:
Set \( p(x) = 0 \)
\( cx + d = 0 \)

\( \implies cx = -d \)

\( \implies x = -\frac{d}{c} \).
Thus, \( x = -\frac{d}{c} \) is the zero of polynomial \( p(x) \).
In simple words: To find the zero of a polynomial, you just need to set the whole polynomial expression equal to zero and then solve that simple equation for the variable, like \( x \). The number you get for \( x \) is the zero.