NCERT Solutions for Class 9 Maths Exercise 2.3 Question 5
Understanding the Question 🧐
This question asks us to factorise four different cubic polynomials. Unlike quadratic polynomials, factorising cubic polynomials requires a few more steps. The main method we’ll use is the Factor Theorem combined with polynomial long division. Let’s break down each part step-by-step!
Factorise:
(i) &&x^3 – 2x^2 – x + 2&&
(ii) &&x^3 – 3x^2 – 9x – 5&&
(iii) &&x^3 + 13x^2 + 32x + 20&&
(iv) &&2y^3 + y^2 – 2y – 1&&
Part (i): Factorise &&x^3 – 2x^2 – x + 2&& 📝
Let’s denote the polynomial as &&p(x) = x^3 – 2x^2 – x + 2&&.
Step 1: Find one factor using the Factor Theorem.
We need to find a value for &&x&& that makes &&p(x) = 0&&. We test the integer factors of the constant term, which is &&2&&. The factors are &&\pm 1&& and &&\pm 2&&.
- Let’s try &&x = 1&&:
&&p(1) = (1)^3 – 2(1)^2 – (1) + 2 = 1 – 2 – 1 + 2 = 0&&.
Since &&p(1) = 0&&, the Factor Theorem tells us that &&(x – 1)&& is a factor of &&p(x)&&.
Step 2: Divide the polynomial by the known factor.
Now, we use polynomial long division to divide &&x^3 – 2x^2 – x + 2&& by &&(x – 1)&&.
On dividing, we get the quotient as &&x^2 – x – 2&&.
Step 3: Factorise the quadratic quotient.
We need to factorise &&x^2 – x – 2&&. We can do this by splitting the middle term.
&&x^2 – x – 2 = x^2 – 2x + x – 2&&
&&= x(x – 2) + 1(x – 2)&&
&&= (x + 1)(x – 2)&&
Step 4: Combine all factors.
The factors of the cubic polynomial are the factor we found in Step 1 and the factors from Step 3.
Therefore, &&x^3 – 2x^2 – x + 2 = \mathbf{(x – 1)(x + 1)(x – 2)}&&.
Part (ii): Factorise &&x^3 – 3x^2 – 9x – 5&& 📝
Let &&p(x) = x^3 – 3x^2 – 9x – 5&&.
Step 1: Find one factor using the Factor Theorem.
The constant term is &&-5&&. Its factors are &&\pm 1&& and &&\pm 5&&.
- Let’s try &&x = 1&&: &&p(1) = 1 – 3 – 9 – 5 = -16 \neq 0&&.
- Let’s try &&x = -1&&:
&&p(-1) = (-1)^3 – 3(-1)^2 – 9(-1) – 5 = -1 – 3 + 9 – 5 = 0&&.
Since &&p(-1) = 0&&, &&(x – (-1)) = (x + 1)&& is a factor of &&p(x)&&.
Step 2: Divide the polynomial by &&(x + 1)&&.
Using long division for &&(x^3 – 3x^2 – 9x – 5) \div (x + 1)&&, we get the quotient &&x^2 – 4x – 5&&.
Step 3: Factorise the quadratic quotient.
We factorise &&x^2 – 4x – 5&& by splitting the middle term.
&&x^2 – 4x – 5 = x^2 – 5x + x – 5&&
&&= x(x – 5) + 1(x – 5)&&
&&= (x + 1)(x – 5)&&
Step 4: Combine all factors.
The complete factorisation is:
&&x^3 – 3x^2 – 9x – 5 = \mathbf{(x + 1)(x + 1)(x – 5)} = \mathbf{(x + 1)^2(x – 5)}&&.
Part (iii): Factorise &&x^3 + 13x^2 + 32x + 20&& 📝
Let &&p(x) = x^3 + 13x^2 + 32x + 20&&.
Step 1: Find one factor.
The constant term is &&20&&. Its factors are &&\pm 1, \pm 2, \pm 4, \pm 5, \dots&&. Since all coefficients are positive, any positive value of &&x&& will result in a positive answer. So, we must test negative factors.
- Let’s try &&x = -1&&:
&&p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0&&.
Since &&p(-1) = 0&&, &&(x + 1)&& is a factor.
Step 2: Divide the polynomial by &&(x + 1)&&.
Dividing &&x^3 + 13x^2 + 32x + 20&& by &&(x + 1)&& gives the quotient &&x^2 + 12x + 20&&.
Step 3: Factorise the quadratic quotient.
We factorise &&x^2 + 12x + 20&&.
&&x^2 + 12x + 20 = x^2 + 10x + 2x + 20&&
&&= x(x + 10) + 2(x + 10)&&
&&= (x + 2)(x + 10)&&
Step 4: Combine all factors.
The complete factorisation is:
&&x^3 + 13x^2 + 32x + 20 = \mathbf{(x + 1)(x + 2)(x + 10)}&&.
Part (iv): Factorise &&2y^3 + y^2 – 2y – 1&& 📝
For this polynomial, let &&p(y) = 2y^3 + y^2 – 2y – 1&&. We can solve this using two methods.
Method 1: Factorisation by Grouping
Sometimes, we can find factors by grouping terms. This is often faster than long division.
&&2y^3 + y^2 – 2y – 1 = (2y^3 + y^2) – (2y + 1)&&
Group the first two and last two terms. Be careful with the negative sign!
&&= y^2(2y + 1) – 1(2y + 1)&&
Factor out the common term from each group.
&&= (y^2 – 1)(2y + 1)&&
We know the identity &&a^2 – b^2 = (a-b)(a+b)&&. Here &&y^2 – 1 = y^2 – 1^2&&.
&&= \mathbf{(y – 1)(y + 1)(2y + 1)}&&.
Method 2: Using the Factor Theorem (for verification)
Step 1: Test factors of the constant term &&-1&&, which are &&\pm 1&&.
- Let’s try &&y = 1&&:
&&p(1) = 2(1)^3 + (1)^2 – 2(1) – 1 = 2 + 1 – 2 – 1 = 0&&.
So, &&(y – 1)&& is a factor.
Step 2: Divide &&2y^3 + y^2 – 2y – 1&& by &&(y – 1)&&.
The division gives a quotient of &&2y^2 + 3y + 1&&.
Step 3: Factorise the quadratic &&2y^2 + 3y + 1&&.
&&2y^2 + 3y + 1 = 2y^2 + 2y + y + 1&&
&&= 2y(y + 1) + 1(y + 1)&&
&&= (2y + 1)(y + 1)&&
Step 4: The final factors are &&(y – 1)(y + 1)(2y + 1)&&, which matches the result from Method 1.
Conclusion and Key Points ✅
We have successfully factorised all four cubic polynomials using the Factor Theorem and polynomial long division. The key is to find one root, divide to get a quadratic, and then factorise the simpler quadratic expression. This ncert solutions guide shows a systematic approach to tackle these problems.
- Factor Theorem: If &&p(a) = 0&& for a polynomial &&p(x)&&, then &&(x – a)&& is a factor of &&p(x)&&.
- Rational Root Theorem: The possible rational roots are of the form &&\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}&&.
- A cubic polynomial will always have three factors (which can be real or complex, and some may be repeated).
- Long division is a reliable method to reduce the degree of the polynomial after finding one root. For some polynomials, grouping might be a faster alternative.
FAQ
Q: What is the first step to factorise a cubic polynomial like &&x^3 – 2x^2 – x + 2&&?
A: The first step is to use the Rational Root Theorem to find a potential root by testing the integer factors of the constant term. For &&x^3 – 2x^2 – x + 2&&, the constant term is &&2&&, so we test the factors &&\pm 1&& and &&\pm 2&&.
Q: How do we confirm if a number is a root of the polynomial?
A: We use the Factor Theorem. If substituting a value ‘&&a&&’ into the polynomial &&p(x)&& results in &&p(a) = 0&&, then &&(x – a)&& is a factor of the polynomial.
Q: What method is used after finding one factor of a cubic polynomial?
A: After finding one linear factor (e.g., &&x – 1&&), we use polynomial long division to divide the cubic polynomial by this factor. This simplifies the problem by giving a quadratic quotient.
Q: How do you factorise the resulting quadratic polynomial?
A: The quadratic polynomial can be factorised by using the ‘splitting the middle term’ method, which involves finding two numbers that sum to the middle coefficient and multiply to the product of the first and last coefficients.
Q: What are the final factors for &&x^3 – 3x^2 – 9x – 5&&?
A: The final factors are &&(x + 1)&&, &&(x + 1)&&, and &&(x – 5)&&. The complete factorisation is &&(x + 1)^2(x – 5)&&.
Q: Can we use grouping to factorise &&2y^3 + y^2 – 2y – 1&&?
A: Yes, for some cubic polynomials like this one, grouping is a much faster method. We can group the terms as &&(2y^3 + y^2) – (2y + 1)&& and then take out common factors to arrive at the solution &&(y-1)(y+1)(2y+1)&&.
Further Reading
For more information and practice problems, you can refer to the official NCERT textbooks available on their website. The ncert solutions provided here are designed to supplement your learning. Visit https://ncert.nic.in/ for official resources.
