NCERT Solutions for Class 9 Maths Exercise 2.4 Question 1
Understanding the Question 🧐
In this problem, we need to determine if the linear polynomial &&x + 1&& is a factor of four different given polynomials. To solve this, we will use a very important tool in algebra called the Factor Theorem.
The Factor Theorem states that a polynomial &&p(x)&& has a factor &&(x – a)&& if and only if &&p(a) = 0&&.
For our question, the factor is &&x + 1&&. We can write this as &&x – (-1)&&. So, here, our ‘&&a&&’ value is &&-1&&. This means we need to check if &&p(-1) = 0&& for each polynomial. If the result is &&0&&, then &&x + 1&& is a factor. If the result is not &&0&&, it is not a factor. Let’s get started with these ncert solutions!
Determine which of the following polynomials has &&(x + 1)&& a factor.
Part (i): &&x^3 + x^2 + x + 1&& 📝
Let the polynomial be &&p(x) = x^3 + x^2 + x + 1&&.
Step 1: Find the zero of the factor.
The factor given is &&x + 1&&. To find its zero, we set it to zero: &&x + 1 = 0 \Rightarrow x = -1&&.
Step 2: Substitute this zero into the polynomial.
Now, we replace every &&x&& in &&p(x)&& with &&-1&&.
&&p(-1) = (-1)^3 + (-1)^2 + (-1) + 1&&
Step 3: Calculate the result.
Remember that &&(-1)&& raised to an odd power is &&-1&&, and &&(-1)&& raised to an even power is &&1&&.
&&p(-1) = -1 + 1 – 1 + 1&&
&&p(-1) = 0&&
Step 4: Conclude based on the Factor Theorem.
Since &&p(-1) = 0&&, the remainder is &&0&&. Therefore, &&(x + 1)&& is a factor of &&x^3 + x^2 + x + 1&&.
Part (ii): &&x^4 + x^3 + x^2 + x + 1&& 📝
Let the polynomial be &&p(x) = x^4 + x^3 + x^2 + x + 1&&.
Step 1: Use the same zero from the factor.
The zero of &&x + 1&& is &&x = -1&&.
Step 2: Substitute &&x = -1&& into the new polynomial.
&&p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1&&
Step 3: Calculate the result.
&&p(-1) = 1 – 1 + 1 – 1 + 1&&
&&p(-1) = 1&&
Step 4: Conclude based on the Factor Theorem.
Since &&p(-1) = 1 \neq 0&&, the remainder is not &&0&&. Therefore, &&(x + 1)&& is not a factor of &&x^4 + x^3 + x^2 + x + 1&&.
Part (iii): &&x^4 + 3x^3 + 3x^2 + x + 1&& 📝
Let the polynomial be &&p(x) = x^4 + 3x^3 + 3x^2 + x + 1&&.
Step 1: Use the zero &&x = -1&&.
Step 2: Substitute &&x = -1&& into the polynomial.
&&p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1&&
Step 3: Calculate the result.
&&p(-1) = 1 + 3(-1) + 3(1) – 1 + 1&&
&&p(-1) = 1 – 3 + 3 – 1 + 1&&
&&p(-1) = 1&&
Step 4: Conclude based on the Factor Theorem.
Since &&p(-1) = 1 \neq 0&&, the remainder is not &&0&&. Therefore, &&(x + 1)&& is not a factor of &&x^4 + 3x^3 + 3x^2 + x + 1&&.
Part (iv): &&x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}&& 📝
Let the polynomial be &&p(x) = x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}&&.
Step 1: Use the zero &&x = -1&&.
Step 2: Substitute &&x = -1&& into the polynomial. Be careful with the signs!
&&p(-1) = (-1)^3 – (-1)^2 – (2 + \sqrt{2})(-1) + \sqrt{2}&&
Step 3: Calculate the result.
&&p(-1) = -1 – 1 – (-(2 + \sqrt{2})) + \sqrt{2}&&
The two negatives in the middle term cancel out, making it positive.
&&p(-1) = -1 – 1 + (2 + \sqrt{2}) + \sqrt{2}&&
&&p(-1) = -2 + 2 + \sqrt{2} + \sqrt{2}&&
&&p(-1) = 0 + 2\sqrt{2}&&
&&p(-1) = 2\sqrt{2}&&
Step 4: Conclude based on the Factor Theorem.
Since &&p(-1) = 2\sqrt{2} \neq 0&&, the remainder is not &&0&&. Therefore, &&(x + 1)&& is not a factor of &&x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}&&.
Conclusion and Key Points ✅
After applying the Factor Theorem to all four polynomials, we found that &&(x + 1)&& is a factor of only the first polynomial, &&x^3 + x^2 + x + 1&&. For the other three polynomials, substituting &&x = -1&& did not result in &&0&&.
- The Factor Theorem is a direct application of the Remainder Theorem.
- To check if &&(x – a)&& is a factor of &&p(x)&&, you must calculate &&p(a)&&.
- If &&p(a) = 0&&, it is a factor.
- If &&p(a) \neq 0&&, it is not a factor.
- The zero of &&(x + 1)&& is &&x = -1&&, and the zero of &&(x – 1)&& is &&x = 1&&.
FAQ
Q: What is the Factor Theorem?
A: The Factor Theorem states that a polynomial &&p(x)&& has a factor &&(x – a)&& if and only if the remainder, which is &&p(a)&&, is equal to &&0&&.
Q: Why do we substitute &&x = -1&& to check for the factor &&(x+1)&&?
A: To find the “zero” of the linear polynomial &&(x+1)&&, we set it equal to zero: &&x + 1 = 0&&. Solving for &&x&& gives &&x = -1&&. The Factor Theorem requires us to test this zero value in the main polynomial.
Q: Is &&(x+1)&& a factor of &&x^3 + x^2 + x + 1&&?
A: Yes. As shown in Part (i), when we substitute &&x = -1&& into the polynomial &&p(x) = x^3 + x^2 + x + 1&&, we get &&p(-1) = 0&&. This confirms that &&(x+1)&& is a factor.
Q: What is the difference between the Remainder Theorem and the Factor Theorem?
A: The Remainder Theorem helps us find the remainder when a polynomial is divided by a linear polynomial like &&(x – a)&&. The Factor Theorem is a special case of this: if the remainder found using the theorem is &&0&&, then &&(x – a)&& is a factor.
Q: How do we handle the term &&-(2 + \sqrt{2})x&& in the fourth polynomial?
A: When you substitute &&x = -1&&, the expression becomes &&-(2 + \sqrt{2})(-1)&&. The two negative signs multiply to become a positive, so the term simplifies to &&+(2 + \sqrt{2})&&.
Further Reading
For more information on polynomials and the Factor Theorem, you can refer to the official NCERT textbook provided by the National Council of Educational Research and Training. Visit NCERT’s official website for textbooks and other resources.
