RBSE Solutions Class 9 Maths Chapter 3 Polynomial Exercise 3.3

Polynomial Ex 3.3

Question 1. Find the remainder when dividing the polynomial \( x^4 + x^3 – 3x^2 + 3x + 1 \) by each of the following one degree expression.
(i) \( x - 1 \)
(ii) \( x - \frac { 1 }{ 2 } \)
(iii) \( x + \pi \)
(iv) \( 3 + 2x \)
(v) \( x \)
Answer:
(i) Let the polynomial be \( p(x) = x^4 + x^3 – 3x^2 + 3x + 1 \).
To find the remainder when divided by \( x - 1 \), we find the root of \( x - 1 = 0 \), which is \( x = 1 \).
Now, substitute \( x = 1 \) into the polynomial \( p(x) \):
\( p(1) = (1)^4 + (1)^3 – 3(1)^2 + 3(1) + 1 \)
\( = 1 + 1 - 3 + 3 + 1 \)
\( = 3 \)
Thus, the remainder when dividing \( p(x) \) by \( x - 1 \) is 3.

(ii) Let the polynomial be \( p(x) = x^4 + x^3 – 3x^2 + 3x + 1 \).
To find the remainder when divided by \( x - \frac { 1 }{ 2 } \), we find the root of \( x - \frac { 1 }{ 2 } = 0 \), which is \( x = \frac { 1 }{ 2 } \).
Now, substitute \( x = \frac { 1 }{ 2 } \) into the polynomial \( p(x) \):
\( p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 1 \)
\( = \frac{1}{16} + \frac{1}{8} - 3\left(\frac{1}{4}\right) + \frac{3}{2} + 1 \)
\( = \frac{1}{16} + \frac{2}{16} - \frac{12}{16} + \frac{24}{16} + \frac{16}{16} \)
\( = \frac{1 + 2 - 12 + 24 + 16}{16} \)
\( = \frac{43 - 12}{16} \)
\( = \frac{31}{16} \)
Therefore, the remainder when dividing \( p(x) \) by \( x - \frac { 1 }{ 2 } \) is \( \frac{31}{16} \).

(iii) Let the polynomial be \( p(x) = x^4 + x^3 – 3x^2 + 3x + 1 \).
To find the remainder when divided by \( x + \pi \), we find the root of \( x + \pi = 0 \), which is \( x = -\pi \).
Now, substitute \( x = -\pi \) into the polynomial \( p(x) \):
\( p(-\pi) = (-\pi)^4 + (-\pi)^3 - 3(-\pi)^2 + 3(-\pi) + 1 \)
\( = \pi^4 - \pi^3 - 3\pi^2 - 3\pi + 1 \)
The remainder is \( \pi^4 - \pi^3 - 3\pi^2 - 3\pi + 1 \).

(iv) Let the polynomial be \( p(x) = x^4 + x^3 – 3x^2 + 3x + 1 \).
To find the remainder when divided by \( 3 + 2x \), we find the root of \( 3 + 2x = 0 \), which is \( 2x = -3 \), so \( x = -\frac{3}{2} \).
Now, substitute \( x = -\frac{3}{2} \) into the polynomial \( p(x) \):
\( p\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^4 + \left(-\frac{3}{2}\right)^3 - 3\left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) + 1 \)
\( = \frac{81}{16} - \frac{27}{8} - 3\left(\frac{9}{4}\right) - \frac{9}{2} + 1 \)
\( = \frac{81}{16} - \frac{54}{16} - \frac{108}{16} - \frac{72}{16} + \frac{16}{16} \)
\( = \frac{81 - 54 - 108 - 72 + 16}{16} \)
\( = \frac{97 - 234}{16} \)
\( = -\frac{137}{16} \)
Thus, the remainder when dividing \( p(x) \) by \( 3 + 2x \) is \( -\frac{137}{16} \).

(v) Let the polynomial be \( p(x) = x^4 + x^3 – 3x^2 + 3x + 1 \).
To find the remainder when divided by \( x \), we find the root of \( x = 0 \), which is \( x = 0 \).
Now, substitute \( x = 0 \) into the polynomial \( p(x) \):
\( p(0) = (0)^4 + (0)^3 - 3(0)^2 + 3(0) + 1 \)
\( = 0 + 0 - 0 + 0 + 1 \)
\( = 1 \)
Therefore, the remainder when dividing \( p(x) \) by \( x \) is 1.
In simple words: To find the remainder when dividing a polynomial by a linear expression, you just need to find the value of x that makes the linear expression zero. Then, put that x value into the polynomial, and the result is your remainder. This is known as the Remainder Theorem.

🎯 Exam Tip: Remember the Remainder Theorem: if a polynomial \( p(x) \) is divided by \( (x-a) \), the remainder is \( p(a) \). This makes finding remainders very quick!

Question 3. Verify whether \( x + 1 \) is a factor of \( x^3 + 3x^2 + 3x + 1 \) or not.
Answer: Let the polynomial be \( p(x) = x^3 + 3x^2 + 3x + 1 \).
For \( x + 1 \) to be a factor of \( p(x) \), the remainder when \( p(x) \) is divided by \( x + 1 \) must be 0.
To find the remainder, we find the root of \( x + 1 = 0 \), which is \( x = -1 \).
Now, substitute \( x = -1 \) into the polynomial \( p(x) \):
\( p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 \)
\( = -1 + 3(1) - 3 + 1 \)
\( = -1 + 3 - 3 + 1 \)
\( = 0 \)
Since \( p(-1) = 0 \), the remainder is 0.
This means \( x + 1 \) is indeed a factor of the polynomial \( x^3 + 3x^2 + 3x + 1 \).
In simple words: To check if \( x + 1 \) is a factor, we see if the polynomial becomes zero when \( x \) is \( -1 \). If it does, then \( x + 1 \) is a factor. Here, it became zero, so it is a factor.

🎯 Exam Tip: The Factor Theorem is a special case of the Remainder Theorem. If \( p(a) = 0 \), then \( (x-a) \) is a factor of \( p(x) \). This is crucial for solving polynomial problems.

Question 4. Polynomial \( x^3 + x^2 - 4x + a \) and \( 2x^3 + ax^2 + 3x – 3 \) when divided by \( x – 2 \) gives same remainder. Find the value of \( a \).
Answer: Let the first polynomial be \( p(x) = x^3 + x^2 - 4x + a \).
When \( p(x) \) is divided by \( x - 2 \), the remainder is \( p(2) \).
Substitute \( x = 2 \) into \( p(x) \):
\( p(2) = (2)^3 + (2)^2 - 4(2) + a \)
\( = 8 + 4 - 8 + a \)
\( = a + 4 \)

Now, let the second polynomial be \( f(x) = 2x^3 + ax^2 + 3x – 3 \).
When \( f(x) \) is divided by \( x - 2 \), the remainder is \( f(2) \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 2(2)^3 + a(2)^2 + 3(2) - 3 \)
\( = 2(8) + a(4) + 6 - 3 \)
\( = 16 + 4a + 6 - 3 \)
\( = 4a + 19 \)

The problem states that both polynomials give the same remainder when divided by \( x - 2 \).
So, we can set the two remainders equal to each other:
\( p(2) = f(2) \)
\( a + 4 = 4a + 19 \)
Now, solve for \( a \):
\( 4 - 19 = 4a - a \)
\( -15 = 3a \)
\( \implies \) \( a = \frac{-15}{3} \)
\( \implies \) \( a = -5 \)
Therefore, the value of \( a \) is -5.
In simple words: We found the remainder for both polynomials by putting \( x=2 \) into them. Since the remainders are equal, we set the two remainder expressions equal to each other and then solved to find the value of \( a \).

🎯 Exam Tip: When two polynomials yield the same remainder upon division by a common factor, setting their remainders equal to each other is the key to solving for unknown coefficients. Remember to substitute the root of the divisor into each polynomial.