GSEB Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1. Find the remainder when \( x^3 + 3x^2 + 3x + 1 \) is divided by
1. \( x + 1 \)
2. \( x - \frac {1}{2} \)
3. \( x \)
4. \( x + \pi \)
5. \( 5 + 2x \)
Answer:
We are using the Remainder Theorem here. To find the remainder when a polynomial \( p(x) \) is divided by \( x - a \), we just need to calculate \( p(a) \). Our polynomial is \( p(x) = x^3 + 3x^2 + 3x + 1 \).
1. When divided by \( x + 1 \):
Set \( x + 1 = 0 \), which gives \( x = -1 \).
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 \)
\( p(-1) = -1 + 3(1) - 3 + 1 \)
\( p(-1) = -1 + 3 - 3 + 1 \)
\( p(-1) = 0 \)
The remainder is 0.
2. When divided by \( x - \frac {1}{2} \):
Set \( x - \frac {1}{2} = 0 \), which gives \( x = \frac {1}{2} \).
Now, substitute \( x = \frac {1}{2} \) into \( p(x) \):
\( p(\frac {1}{2}) = (\frac {1}{2})^3 + 3(\frac {1}{2})^2 + 3(\frac {1}{2}) + 1 \)
\( p(\frac {1}{2}) = \frac {1}{8} + 3(\frac {1}{4}) + \frac {3}{2} + 1 \)
\( p(\frac {1}{2}) = \frac {1}{8} + \frac {3}{4} + \frac {3}{2} + 1 \)
To add these fractions, we find a common denominator, which is 8.
\( p(\frac {1}{2}) = \frac {1}{8} + \frac {3 \times 2}{4 \times 2} + \frac {3 \times 4}{2 \times 4} + \frac {1 \times 8}{1 \times 8} \)
\( p(\frac {1}{2}) = \frac {1}{8} + \frac {6}{8} + \frac {12}{8} + \frac {8}{8} \)
\( p(\frac {1}{2}) = \frac {1 + 6 + 12 + 8}{8} \)
\( p(\frac {1}{2}) = \frac {27}{8} \)
The remainder is \( \frac {27}{8} \).
3. When divided by \( x \):
Set \( x = 0 \).
Now, substitute \( x = 0 \) into \( p(x) \):
\( p(0) = (0)^3 + 3(0)^2 + 3(0) + 1 \)
\( p(0) = 0 + 0 + 0 + 1 \)
\( p(0) = 1 \)
The remainder is 1.
4. When divided by \( x + \pi \):
Set \( x + \pi = 0 \), which gives \( x = -\pi \).
Now, substitute \( x = -\pi \) into \( p(x) \):
\( p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1 \)
\( p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1 \)
The remainder is \( -\pi^3 + 3\pi^2 - 3\pi + 1 \).
5. When divided by \( 5 + 2x \):
Set \( 5 + 2x = 0 \), which gives \( 2x = -5 \), so \( x = -\frac {5}{2} \).
Now, substitute \( x = -\frac {5}{2} \) into \( p(x) \):
\( p(-\frac {5}{2}) = (-\frac {5}{2})^3 + 3(-\frac {5}{2})^2 + 3(-\frac {5}{2}) + 1 \)
\( p(-\frac {5}{2}) = -\frac {125}{8} + 3(\frac {25}{4}) - \frac {15}{2} + 1 \)
\( p(-\frac {5}{2}) = -\frac {125}{8} + \frac {75}{4} - \frac {15}{2} + 1 \)
To add and subtract these fractions, we find a common denominator, which is 8.
\( p(-\frac {5}{2}) = -\frac {125}{8} + \frac {75 \times 2}{4 \times 2} - \frac {15 \times 4}{2 \times 4} + \frac {1 \times 8}{1 \times 8} \)
\( p(-\frac {5}{2}) = -\frac {125}{8} + \frac {150}{8} - \frac {60}{8} + \frac {8}{8} \)
\( p(-\frac {5}{2}) = \frac {-125 + 150 - 60 + 8}{8} \)
\( p(-\frac {5}{2}) = \frac {25 - 60 + 8}{8} \)
\( p(-\frac {5}{2}) = \frac {-35 + 8}{8} \)
\( p(-\frac {5}{2}) = \frac {-27}{8} \)
The remainder is \( -\frac {27}{8} \).
In simple words: To find the remainder quickly, set the part you are dividing by to zero to find the value of \( x \). Then, put that value of \( x \) into the main polynomial and calculate the result. That number is your remainder.

Exam Tip: Remember to simplify fractions and combine terms carefully, especially when dealing with negative numbers or fractions, to avoid calculation errors.

Question 2. Find the remainder when \( x^3 - ax^2 + 6x - a \) is divided by \( x - a \).
Answer:
We will use the Remainder Theorem. The polynomial is \( p(x) = x^3 - ax^2 + 6x - a \).
The divisor is \( x - a \). To find the remainder, we set the divisor to zero:
\( x - a = 0 \)
\( \implies x = a \)
Now, substitute this value of \( x \) into the polynomial \( p(x) \):
\( p(a) = (a)^3 - a(a)^2 + 6(a) - a \)
\( p(a) = a^3 - a^3 + 6a - a \)
\( p(a) = 0 + 5a \)
\( p(a) = 5a \)
Therefore, the remainder when \( x^3 - ax^2 + 6x - a \) is divided by \( x - a \) is \( 5a \).
In simple words: To find the remainder when a polynomial is divided by \( x \) minus some number (or variable), just put that number (or variable) into the polynomial instead of \( x \) and calculate the result. This result will be the remainder.

Exam Tip: The Remainder Theorem is a fundamental concept. Ensure you substitute the value correctly, especially when dealing with variables like 'a' in the polynomial terms.

Question 3. Check whether \( 7 + 3x \) is a factor of \( 3x^3 + 7x \).
Answer:
For \( 7 + 3x \) to be a factor of the polynomial \( p(x) = 3x^3 + 7x \), the remainder when \( p(x) \) is divided by \( 7 + 3x \) must be 0.
First, we find the value of \( x \) by setting the potential factor to zero:
\( 7 + 3x = 0 \)
\( \implies 3x = -7 \)
\( \implies x = -\frac {7}{3} \)
Now, substitute this value of \( x \) into the polynomial \( p(x) \):
\( p(-\frac {7}{3}) = 3(-\frac {7}{3})^3 + 7(-\frac {7}{3}) \)
\( p(-\frac {7}{3}) = 3(-\frac {343}{27}) - \frac {49}{3} \)
\( p(-\frac {7}{3}) = -\frac {343}{9} - \frac {49}{3} \)
To subtract these fractions, we find a common denominator, which is 9.
\( p(-\frac {7}{3}) = -\frac {343}{9} - \frac {49 \times 3}{3 \times 3} \)
\( p(-\frac {7}{3}) = -\frac {343}{9} - \frac {147}{9} \)
\( p(-\frac {7}{3}) = \frac {-343 - 147}{9} \)
\( p(-\frac {7}{3}) = \frac {-490}{9} \)
Since the remainder \( (-\frac {490}{9}) \) is not equal to 0, \( 7 + 3x \) is not a factor of \( 3x^3 + 7x \).
In simple words: To see if one expression is a factor of another, set the first expression to zero to find \( x \). Then, put that \( x \) into the second expression. If the answer is zero, it's a factor; if not, it's not a factor. In this case, it wasn't zero, so it's not a factor.

Exam Tip: Remember the Factor Theorem: a polynomial \( p(x) \) has \( (x - a) \) as a factor if and only if \( p(a) = 0 \). Always ensure your final calculation for the remainder is accurate.