NCERT Solutions for Class 9 Maths Exercise 2.4 Question 5
Understanding the Question 🧐
This question requires us to factorise cubic polynomials. Unlike quadratic polynomials, there isn’t a simple formula at this level. We will use a reliable multi-step method. These ncert solutions will break down the process.
The Four-Step Method:
- Hit and Trial: Find one root of the polynomial by testing the integer factors of the constant term. If &&p(a) = 0&&, then &&a&& is a root.
- Find the First Factor: If &&a&& is a root, then &&(x – a)&& is a factor of &&p(x)&&, according to the Factor Theorem.
- Polynomial Long Division: Divide the cubic polynomial by the linear factor you just found. The result will be a quadratic polynomial.
- Factorise the Quadratic: Factor the resulting quadratic quotient using the “splitting the middle term” method. The final answer will be the product of three linear factors.
Factorise the following:
Part (i): &&x^3 – 2x^2 – x + 2&& 📝
Step 1: Hit and Trial.
Let &&p(x) = x^3 – 2x^2 – x + 2&&. The constant term is &&2&&. The factors of &&2&& are &&\pm 1, \pm 2&&. Let’s test &&x=1&&.
&&p(1) = (1)^3 – 2(1)^2 – (1) + 2 = 1 – 2 – 1 + 2 = 0&&.
Since &&p(1)=0&&, we’ve found a root!
Step 2: Find the First Factor.
Because &&x=1&& is a root, &&(x-1)&& is a factor of &&p(x)&&.
Step 3: Polynomial Long Division.
We divide &&(x^3 – 2x^2 – x + 2)&& by &&(x-1)&&. The division gives a quotient of &&x^2 – x – 2&&.
Step 4: Factorise the Quadratic.
Now, we factorise &&x^2 – x – 2&& by splitting the middle term. We need two numbers that add to &&-1&& and multiply to &&-2&&. The numbers are &&-2&& and &&1&&.
&&x^2 – 2x + x – 2 = x(x-2) + 1(x-2) = (x-2)(x+1)&&.
The final factors are the combination of the factor from Step 2 and the factors from Step 4.
Final Answer: &&(x-1)(x+1)(x-2)&&
Alternative Method: This particular polynomial can also be factored by grouping: &&x^2(x-2) – 1(x-2) = (x^2-1)(x-2) = (x-1)(x+1)(x-2)&&.
Part (ii): &&x^3 – 3x^2 – 9x – 5&& 📝
Step 1: Hit and Trial.
Let &&p(x) = x^3 – 3x^2 – 9x – 5&&. The constant term is &&-5&&. Factors are &&\pm 1, \pm 5&&. Let’s test &&x=-1&&.
&&p(-1) = (-1)^3 – 3(-1)^2 – 9(-1) – 5 = -1 – 3(1) + 9 – 5 = -1 – 3 + 9 – 5 = 0&&.
Step 2: Find the First Factor.
Since &&x=-1&& is a root, &&(x – (-1)) = (x+1)&& is a factor.
Step 3: Polynomial Long Division.
Divide &&(x^3 – 3x^2 – 9x – 5)&& by &&(x+1)&&. The quotient is &&x^2 – 4x – 5&&.
Step 4: Factorise the Quadratic.
Factorise &&x^2 – 4x – 5&&. We need numbers that sum to &&-4&& and multiply to &&-5&&. These are &&-5&& and &&1&&.
&&x^2 – 5x + x – 5 = x(x-5) + 1(x-5) = (x-5)(x+1)&&.
Final Answer: &&(x+1)(x+1)(x-5) = (x+1)^2(x-5)&&
Part (iii): &&x^3 + 13x^2 + 32x + 20&& 📝
Step 1: Hit and Trial.
Let &&p(x) = x^3 + 13x^2 + 32x + 20&&. Factors of &&20&& are &&\pm 1, \pm 2, \ldots&&. Since all coefficients are positive, the root must be negative. Let’s try &&x=-1&&.
&&p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0&&.
Step 2: Find the First Factor.
Since &&x=-1&& is a root, &&(x+1)&& is a factor.
Step 3: Polynomial Long Division.
Divide &&(x^3 + 13x^2 + 32x + 20)&& by &&(x+1)&&. The quotient is &&x^2 + 12x + 20&&.
Step 4: Factorise the Quadratic.
Factorise &&x^2 + 12x + 20&&. We need numbers that sum to &&12&& and multiply to &&20&&. These are &&10&& and &&2&&.
&&x^2 + 10x + 2x + 20 = x(x+10) + 2(x+10) = (x+10)(x+2)&&.
Final Answer: &&(x+1)(x+2)(x+10)&&
Part (iv): &&2y^3 + y^2 – 2y – 1&& 📝
Step 1: Hit and Trial.
Let &&p(y) = 2y^3 + y^2 – 2y – 1&&. The constant term is &&-1&&. Factors are &&\pm 1&&. Let’s try &&y=1&&.
&&p(1) = 2(1)^3 + (1)^2 – 2(1) – 1 = 2 + 1 – 2 – 1 = 0&&.
Step 2: Find the First Factor.
Since &&y=1&& is a root, &&(y-1)&& is a factor.
Step 3: Polynomial Long Division.
Divide &&(2y^3 + y^2 – 2y – 1)&& by &&(y-1)&&. The quotient is &&2y^2 + 3y + 1&&.
Step 4: Factorise the Quadratic.
Factorise &&2y^2 + 3y + 1&&. We need numbers that sum to &&3&& and multiply to &&2 \times 1 = 2&&. These are &&2&& and &&1&&.
&&2y^2 + 2y + y + 1 = 2y(y+1) + 1(y+1) = (y+1)(2y+1)&&.
Final Answer: &&(y-1)(y+1)(2y+1)&&
Alternative Method: This polynomial can also be factored by grouping: &&y^2(2y+1) – 1(2y+1) = (y^2-1)(2y+1) = (y-1)(y+1)(2y+1)&&.
Conclusion and Key Points ✅
Factorising cubic polynomials is a systematic process that combines the Factor Theorem, polynomial division, and quadratic factorisation. By mastering these four steps, you can break down complex cubic expressions into their simpler linear factors.
- Find a root: Use the “Hit and Trial” method with factors of the constant term.
- Find a factor: If &&p(a)=0&&, then &&(x-a)&& is a factor.
- Divide: Use long division to divide the cubic by the linear factor to get a quadratic.
- Factorise: Factor the resulting quadratic to find the other two linear factors.
FAQ
Q: What is the first step to factorise a cubic polynomial?
A: The first step is to find one linear factor using the ‘Hit and Trial’ method. Test the integer factors of the constant term to find a value ‘&&a&&’ for which &&p(a) = 0&&. Once found, &&(x-a)&& is your first factor.
Q: What do you do after finding one factor of a cubic polynomial?
A: After finding one linear factor, you use polynomial long division to divide the original cubic polynomial by that factor. The result (the quotient) will be a quadratic polynomial, which you can then factorise separately.
Q: How do you factorise &&x^3 – 2x^2 – x + 2&&?
A: First, find a root by trial. &&p(1) = 1 – 2 – 1 + 2 = 0&&, so &&(x-1)&& is a factor. Divide the polynomial by &&(x-1)&& to get the quadratic &&x^2 – x – 2&&. Factor this quadratic to get &&(x-2)(x+1)&&. The final factors are &&(x-1)(x+1)(x-2)&&.
Q: Is there a shortcut instead of long division?
A: Sometimes, yes. If the polynomial has the right structure of coefficients, you can factor it by grouping terms, which is much faster. A method called synthetic division is also quicker but is not always taught in the standard NCERT curriculum.
Q: How does the ‘Hit and Trial’ method work?
A: You list all the integer factors (both positive and negative) of the constant term of the polynomial. Then you substitute these factors one by one into the polynomial until you find one that makes the polynomial equal to zero. This value is a root of the polynomial.
Further Reading
For more practice and a deeper understanding of factorising polynomials, refer to Chapter 2 of your Class 9 Maths textbook. The official NCERT books are available on their website: https://ncert.nic.in/.
