NCERT Solutions for Class 9 Maths Exercise 2.3 Question 1
Understanding the Question 🧐
This question asks us to find the remainder when the polynomial &&p(x) = x^3 + 3x^2 + 3x + 1&& is divided by several different linear polynomials. Instead of using long division, we can use a simple and powerful shortcut called the Remainder Theorem.
To use this theorem, we will follow two simple steps for each part:
- Find the zero of the divisor (the polynomial we are dividing by).
- Substitute this zero into the main polynomial &&p(x)&& to find the remainder.
Find the remainder when &&x^3 + 3x^2 + 3x + 1&& is divided by:
(i) &&x + 1&& (ii) &&x – \frac{1}{2}&& (iii) &&x&& (iv) &&x + \pi&& (v) &&5 + 2x&&
Part (i): Divided by &&x + 1&& 📝
Step 1: Find the zero of the divisor &&x+1&&.
Set &&x + 1 = 0&&. This gives &&x = -1&&.
Step 2: Substitute &&x = -1&& into &&p(x)&&.
&&p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1&&
&&= -1 + 3(1) – 3 + 1&&
&&= -1 + 3 – 3 + 1 = 0&&
The remainder is 0.
Part (ii): Divided by &&x – \frac{1}{2}&& 📝
Step 1: Find the zero of the divisor &&x – \frac{1}{2}&&.
Set &&x – \frac{1}{2} = 0&&. This gives &&x = \frac{1}{2}&&.
Step 2: 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&&
&&= \frac{1}{8} + 3(\frac{1}{4}) + \frac{3}{2} + 1&&
&&= \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1&&
To add these, find a common denominator (8): &&= \frac{1}{8} + \frac{6}{8} + \frac{12}{8} + \frac{8}{8} = \frac{1+6+12+8}{8} = \frac{27}{8}&&
The remainder is &&\frac{27}{8}&&.
Part (iii): Divided by &&x&& 📝
Step 1: Find the zero of the divisor &&x&&.
Set &&x = 0&&.
Step 2: Substitute &&x = 0&& into &&p(x)&&.
&&p(0) = (0)^3 + 3(0)^2 + 3(0) + 1 = 0 + 0 + 0 + 1 = 1&&
The remainder is 1.
Part (iv): Divided by &&x + \pi&& 📝
Step 1: Find the zero of the divisor &&x + \pi&&.
Set &&x + \pi = 0&&. This gives &&x = -\pi&&.
Step 2: Substitute &&x = -\pi&& into &&p(x)&&.
&&p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1&&
&&= -\pi^3 + 3\pi^2 – 3\pi + 1&&
This expression cannot be simplified further.
The remainder is &&-\pi^3 + 3\pi^2 – 3\pi + 1&&.
Part (v): Divided by &&5 + 2x&& 📝
Step 1: Find the zero of the divisor &&5 + 2x&&.
Set &&5 + 2x = 0&&. This gives &&2x = -5&&, so &&x = -\frac{5}{2}&&.
Step 2: 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&&
&&= -\frac{125}{8} + 3(\frac{25}{4}) – \frac{15}{2} + 1&&
&&= -\frac{125}{8} + \frac{75}{4} – \frac{15}{2} + 1&&
Find a common denominator (8): &&= -\frac{125}{8} + \frac{150}{8} – \frac{60}{8} + \frac{8}{8} = \frac{-125+150-60+8}{8} = \frac{25-60+8}{8} = \frac{-27}{8}&&
The remainder is &&-\frac{27}{8}&&.
FAQ ❓
Q: What is the Remainder Theorem?
A: The Remainder Theorem is a shortcut in polynomial algebra. It states that if you divide a polynomial &&p(x)&& by a linear factor &&(x – a)&&, the remainder of that division is simply the value you get when you calculate &&p(a)&&.
Q: What happens if the remainder is 0?
A: If the remainder is 0, it means the divisor is a perfect factor of the polynomial. In Part (i), since the remainder was 0, we know that &&(x+1)&& is a factor of &&x^3 + 3x^2 + 3x + 1&&.
Further Reading 📖
To learn more about the Remainder Theorem and polynomial division, you can refer to the official NCERT textbook for Class 9 Maths, Chapter 2. More resources are available on the NCERT website at https://ncert.nic.in/.
