NCERT Solutions for Class 9 Maths Exercise 2.4 Question 2
Understanding the Question 🧐
This question asks us to use the Factor Theorem to check if a polynomial, &&g(x)&&, is a factor of another polynomial, &&p(x)&&. This is a direct application of the theorem we’ve been learning about.
The process is simple:
- Find the zero of &&g(x)&&.
- Substitute this zero into &&p(x)&&.
- If the result is &&0&&, then &&g(x)&& is a factor. Otherwise, it is not.
Use the Factor Theorem to determine whether &&g(x)&& is a factor of &&p(x)&& in each of the following cases.
Part (i): &&p(x) = 2x^3 + x^2 – 2x – 1&&, &&g(x) = x + 1&& 📝
Step 1: Find the zero of &&g(x)&&.
To find the zero of &&g(x) = x + 1&&, we set it equal to zero and solve for &&x&&.
&&x + 1 = 0 \Rightarrow x = -1&&.
Step 2: Substitute this zero into &&p(x)&&.
Now, we calculate &&p(-1)&& by replacing every &&x&& in &&p(x)&& with &&-1&&.
&&p(-1) = 2(-1)^3 + (-1)^2 – 2(-1) – 1&&
Step 3: Calculate the result.
&&p(-1) = 2(-1) + (1) – (-2) – 1&&
&&p(-1) = -2 + 1 + 2 – 1&&
&&p(-1) = 0&&
Step 4: Conclusion.
Since &&p(-1) = 0&&, the Factor Theorem tells us that &&g(x) = x + 1&& is a factor of &&p(x) = 2x^3 + x^2 – 2x – 1&&.
Part (ii): &&p(x) = x^3 + 3x^2 + 3x + 1&&, &&g(x) = x + 2&& 📝
Step 1: Find the zero of &&g(x)&&.
For &&g(x) = x + 2&&, we solve &&x + 2 = 0&&, which gives &&x = -2&&.
Step 2: Substitute &&x = -2&& into &&p(x)&&.
&&p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1&&
Step 3: Calculate the result.
&&p(-2) = -8 + 3(4) – 6 + 1&&
&&p(-2) = -8 + 12 – 6 + 1&&
&&p(-2) = 4 – 6 + 1&&
&&p(-2) = -1&&
Step 4: Conclusion.
Since &&p(-2) = -1 \neq 0&&, &&g(x) = x + 2&& is not a factor of &&p(x) = x^3 + 3x^2 + 3x + 1&&.
Part (iii): &&p(x) = x^3 – 4x^2 + x + 6&&, &&g(x) = x – 3&& 📝
Step 1: Find the zero of &&g(x)&&.
For &&g(x) = x – 3&&, we solve &&x – 3 = 0&&, which gives &&x = 3&&.
Step 2: Substitute &&x = 3&& into &&p(x)&&.
&&p(3) = (3)^3 – 4(3)^2 + (3) + 6&&
Step 3: Calculate the result.
&&p(3) = 27 – 4(9) + 3 + 6&&
&&p(3) = 27 – 36 + 3 + 6&&
&&p(3) = -9 + 9&&
&&p(3) = 0&&
Step 4: Conclusion.
Since &&p(3) = 0&&, the Factor Theorem confirms that &&g(x) = x – 3&& is a factor of &&p(x) = x^3 – 4x^2 + x + 6&&.
Conclusion and Key Points ✅
By applying the Factor Theorem, we determined that:
- For case (i), &&g(x)&& is a factor of &&p(x)&&.
- For case (ii), &&g(x)&& is not a factor of &&p(x)&&.
- For case (iii), &&g(x)&& is a factor of &&p(x)&&.
- The Factor Theorem is your main tool for this type of problem.
- Step 1: Find the zero of the linear factor &&g(x)&&.
- Step 2: Substitute this zero value into the main polynomial &&p(x)&&.
- Step 3: If the result is &&0&&, &&g(x)&& is a factor. If the result is anything else, it is not.
FAQ
Q: How do you find the zero of a linear polynomial like &&g(x) = x + 2&&?
A: To find the zero, you set the polynomial equal to zero and solve for the variable. For &&g(x) = x + 2&&, we solve the equation &&x + 2 = 0&&, which gives &&x = -2&&.
Q: What does it mean if &&p(a) \neq 0&& when testing a factor &&(x-a)&&?
A: If &&p(a) \neq 0&&, it means there is a non-zero remainder when &&p(x)&& is divided by &&(x-a)&&. According to the Factor Theorem, this means that &&(x-a)&& is not a factor of &&p(x)&&.
Q: Is &&g(x) = x + 1&& a factor of &&p(x) = 2x^3 + x^2 – 2x – 1&&?
A: Yes. The zero of &&g(x)&& is &&x = -1&&. When we calculate &&p(-1)&&, the result is &&0&&. This confirms that &&g(x)&& is a factor of &&p(x)&&.
Q: Is &&g(x) = x – 3&& a factor of &&p(x) = x^3 – 4x^2 + x + 6&&?
A: Yes. The zero of &&g(x)&& is &&x = 3&&. When we calculate &&p(3)&&, the result is &&0&&. According to the Factor Theorem, &&g(x)&& is a factor of &&p(x)&&.
Q: Can the Factor Theorem be used for quadratic factors like &&x^2 – 1&&?
A: The Factor Theorem directly applies to linear factors of the form &&(x-a)&&. For a quadratic factor like &&x^2 – 1&&, you would first factor it into its linear factors, which are &&(x-1)&& and &&(x+1)&&, and then test each one individually using the theorem.
Further Reading
To deepen your understanding of polynomials, the Factor Theorem, and the Remainder Theorem, consult your Class 9 Maths textbook. The official NCERT textbooks are available on their website at https://ncert.nic.in/.
