NCERT Solutions for Class 9 Maths Chapter 2 Polynomials — Complete Guide with All Exercises

The zero polynomial is the constant polynomial $p(x) = 0$. Its degree is undefined (or sometimes stated as $-\infty$ in advanced texts). The NCERT textbook says the degree of the zero polynomial is "not defined."

Do not confuse the zero polynomial ($p(x) = 0$ for all $x$) with the zeroes of a polynomial (the values of $x$ where $p(x) = 0$). These are completely different concepts.

Question: Which of the following expressions are polynomials in one variable? Which are not? Give reasons.
(i) $4x^2 - 3x + 7$
(ii) $y^2 + \sqrt{2}$
(iii) $3\sqrt{t} + t\sqrt{2}$
(iv) $y + \frac{2}{y}$
(v) $x^{10} + y^3 + t^{50}$

Solution:

**(i) $4x^2 - 3x + 7$ — Yes**, this is a polynomial in $x$. All exponents of $x$ are whole numbers (2, 1, 0).

**(ii) $y^2 + \sqrt{2}$ — Yes**, this is a polynomial in $y$. The $\sqrt{2}$ is a constant coefficient, not a variable with a fractional exponent. Degree = 2.

**(iii) $3\sqrt{t} + t\sqrt{2} = 3t^{1/2} + \sqrt{2}\,t$ — No**, this is NOT a polynomial because the term $3t^{1/2}$ has a fractional exponent ($\frac{1}{2}$). Polynomial exponents must be non-negative integers.

**(iv) $y + \frac{2}{y} = y + 2y^{-1}$ — No**, this is NOT a polynomial because $2y^{-1}$ has a negative exponent.

**(v) $x^{10} + y^3 + t^{50}$ — This is a polynomial, but it is a polynomial in three variables** ($x$, $y$, $t$), not in one variable.

Key Takeaway: For a polynomial in one variable, every exponent of that variable must be a whole number (0, 1, 2, 3, ...).

Question: Write the degree of each of the following polynomials:
(i) $5x^3 + 4x^2 + 7x$
(ii) $4 - y^2$
(iii) $5t - \sqrt{7}$
(iv) $3$

Solution:

(i) Highest power of $x$ is 3. Degree = 3 (cubic polynomial).

(ii) Highest power of $y$ is 2. Degree = 2 (quadratic polynomial).

(iii) Highest power of $t$ is 1. Degree = 1 (linear polynomial).

(iv) $3 = 3x^0$. Highest power of $x$ is 0. Degree = 0 (constant polynomial).

Common Error: Students sometimes say the degree of $3$ is "no degree" or "undefined." Only the zero polynomial has an undefined degree. The constant $3$ has degree 0.

Question: Classify the following as linear, quadratic, and cubic polynomials:
(i) $x^2 + x$
(ii) $x - x^3$
(iii) $y + y^2 + 4$
(iv) $1 + x$
(v) $3t$
(vi) $r^2$
(vii) $7x^3$

Solution:

Question: Find the value of the polynomial $p(x) = 5x - 4x^2 + 3$ at (i) $x = 0$ and (ii) $x = -1$.

Solution:

(i) $p(0) = 5(0) - 4(0)^2 + 3 = 0 - 0 + 3 = 3$

(ii) $p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = -6$

Key Technique: Substitute the value carefully, paying special attention to negative signs. A common error is computing $(-1)^2$ as $-1$ instead of $+1$.

Question: Verify whether the following are zeroes of the polynomial indicated against them:
(i) $p(x) = 3x + 1$, $x = -\frac{1}{3}$
(ii) $p(x) = 5x - \pi$, $x = \frac{4}{5}$
(iii) $p(x) = x^2 - 1$, $x = 1$, $x = -1$
(iv) $p(x) = (x+1)(x-2)$, $x = -1$, $x = 2$
(v) $p(x) = x^2$, $x = 0$

Solution:

(i) $p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1 = -1 + 1 = 0$ ✓ Yes, it is a zero.

(ii) $p\left(\frac{4}{5}\right) = 5 \times \frac{4}{5} - \pi = 4 - \pi \neq 0$ (since $\pi \approx 3.14$). Not a zero.

(iii) $p(1) = 1^2 - 1 = 0$ ✓; $p(-1) = (-1)^2 - 1 = 1 - 1 = 0$ ✓. Both are zeroes.

(iv) $p(-1) = (-1+1)(-1-2) = 0 \times (-3) = 0$ ✓; $p(2) = (2+1)(2-2) = 3 \times 0 = 0$ ✓. Both are zeroes.

(v) $p(0) = 0^2 = 0$ ✓. **Yes, $x = 0$ is a zero** (with multiplicity 2).

Question: Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
(ii) $p(x) = x - 5$
(iii) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax$, $a \neq 0$
(vii) $p(x) = cx + d$, $c \neq 0$

Solution:
For a linear polynomial $p(x) = ax + b$, the zero is $x = -\frac{b}{a}$.

The Factor Theorem is a special case of the Remainder Theorem.

Statement: $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.

Proof: By the Remainder Theorem, the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$. If $p(a) = 0$, then:

The Factor Theorem is your primary tool for factorising cubic (and higher-degree) polynomials. The strategy is:
1. Guess a root $a$ by trying small integers ($0, \pm 1, \pm 2, \ldots$).
2. Confirm that $p(a) = 0$.
3. Divide $p(x)$ by $(x - a)$ to get the remaining quadratic.
4. Factorise the quadratic.

Question: 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) $5 + 2x$

Solution:

(i) Divisor: $(x + 1) = (x - (-1))$, so $a = -1$.

(ii) Divisor: $\left(x - \frac{1}{2}\right)$, so $a = \frac{1}{2}$.

(iii) Divisor: $x = (x - 0)$, so $a = 0$.

(iv) Divisor: $(x + \pi) = (x - (-\pi))$, so $a = -\pi$.

Notice this equals $(1 - \pi)^3$ by the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

(v) Divisor: $(5 + 2x) = 2\left(x + \frac{5}{2}\right) = 2\left(x - \left(-\frac{5}{2}\right)\right)$, so $a = -\frac{5}{2}$.

Question: Find the remainder when $x^3 - ax^2 + 6x - a$ is divided by $(x - a)$.

Solution:
By the Remainder Theorem, remainder $= p(a)$.

**Remainder = $5a$.**

Key Insight: Even when the polynomial has parameters (like $a$), the Remainder Theorem works the same way. Just substitute and simplify.

Question: Check whether $7 + 3x$ is a factor of $3x^3 + 7x$.

Solution:
$(7 + 3x) = 3\left(x + \frac{7}{3}\right)$, so the zero is $x = -\frac{7}{3}$.

Since $p\left(-\frac{7}{3}\right) \neq 0$, $(7 + 3x)$ is not a factor of $3x^3 + 7x$.

Question: Determine which of the following polynomials has $(x + 1)$ as a factor:
(i) $x^3 + x^2 + x + 1$
(ii) $x^4 + x^3 + x^2 + x + 1$
(iii) $x^4 + 3x^3 + 3x^2 + x + 1$
(iv) $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

Solution:
For $(x+1)$ to be a factor, we need $p(-1) = 0$.

(i) $p(-1) = -1 + 1 - 1 + 1 = 0$ ✓ Yes.

(ii) $p(-1) = 1 - 1 + 1 - 1 + 1 = 1 \neq 0$. No.

(iii) $p(-1) = 1 - 3 + 3 - 1 + 1 = 1 \neq 0$. No.

(iv) $p(-1) = -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$. No.

Question: Use the Factor Theorem to factorise $x^3 - 2x^2 - x + 2$.

Solution:
Let $p(x) = x^3 - 2x^2 - x + 2$.

Step 1: Find a root. Constant term = 2, so try $x = \pm 1, \pm 2$.

$p(1) = 1 - 2 - 1 + 2 = 0$ ✓

So $(x - 1)$ is a factor.

**Step 2: Divide by $(x - 1)$.**
Using synthetic division or long division:

Step 3: Factorise the quadratic.
$x^2 - x - 2$: we need two numbers that multiply to $-2$ and add to $-1$. Those are $-2$ and $+1$.

Final Answer:

Verification: The zeroes are $x = 1, 2, -1$. Check: $p(2) = 8 - 8 - 2 + 2 = 0$ ✓, $p(-1) = -1 - 2 + 1 + 2 = 0$ ✓.

Question: Factorise $x^3 - 3x^2 - 9x - 5$.

Solution:
Let $p(x) = x^3 - 3x^2 - 9x - 5$. Constant term = $-5$, so try $x = \pm 1, \pm 5$.

$p(-1) = -1 - 3 + 9 - 5 = 0$ ✓

So $(x + 1)$ is a factor. Dividing:

Factorise $x^2 - 4x - 5$: numbers that multiply to $-5$ and add to $-4$ are $-5$ and $+1$.

Final Answer:

Note: $x = -1$ is a repeated root (multiplicity 2).

Question: Factorise $x^3 + 13x^2 + 32x + 20$.

Solution:
Let $p(x) = x^3 + 13x^2 + 32x + 20$. Constant term = 20, so try factors of 20.

$p(-1) = -1 + 13 - 32 + 20 = 0$ ✓

So $(x + 1)$ is a factor. Dividing:

Factorise $x^2 + 12x + 20$: numbers that multiply to 20 and add to 12 are 10 and 2.

Final Answer:

Question: Factorise $2y^3 + y^2 - 2y - 1$.

Solution:
Let $p(y) = 2y^3 + y^2 - 2y - 1$. Leading coefficient = 2, constant = $-1$.

Possible rational roots: $\pm 1, \pm \frac{1}{2}$.

$p(1) = 2 + 1 - 2 - 1 = 0$ ✓

So $(y - 1)$ is a factor. Dividing:

Factorise $2y^2 + 3y + 1$: we need numbers that multiply to $2 \times 1 = 2$ and add to 3. Those are 2 and 1.

Final Answer:

Question: Use suitable identities to find the following products:
(i) $(x + 4)(x + 10)$
(ii) $(x + 8)(x - 10)$
(iii) $(3x + 4)(3x - 5)$
(iv) $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$
(v) $(3 - 2x)(3 + 2x)$

Solution:

(i) Using Identity IV: $(x + a)(x + b) = x^2 + (a+b)x + ab$

(ii) $(x + 8)(x - 10) = x^2 + (8 + (-10))x + 8 \times (-10) = x^2 - 2x - 80$

(iii) Let $y = 3x$: $(y + 4)(y - 5) = y^2 + (4-5)y + 4(-5) = y^2 - y - 20$

(iv) Using Identity III: $a^2 - b^2 = (a+b)(a-b)$

(v) $(3 - 2x)(3 + 2x) = 3^2 - (2x)^2 = 9 - 4x^2$

Question: Evaluate the following using suitable identities:
(i) $103^2$
(ii) $99^2$
(iii) $105 \times 95$
(iv) $9.8 \times 10.2$

Solution:

(i) $103^2 = (100 + 3)^2 = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609$

(ii) $99^2 = (100 - 1)^2 = 100^2 - 2(100)(1) + 1^2 = 10000 - 200 + 1 = 9801$

(iii) $105 \times 95 = (100 + 5)(100 - 5) = 100^2 - 5^2 = 10000 - 25 = 9975$

(iv) $9.8 \times 10.2 = (10 - 0.2)(10 + 0.2) = 10^2 - 0.2^2 = 100 - 0.04 = 99.96$

Question: Factorise the following:
(i) $4x^2 + 20x + 25$
(ii) $9y^2 - 66yz + 121z^2$
(iii) $(2a - 3b)^2 - (3a - 2b)^2$

Solution:

(i) $4x^2 + 20x + 25 = (2x)^2 + 2(2x)(5) + 5^2 = (2x + 5)^2$ [Identity I]

(ii) $9y^2 - 66yz + 121z^2 = (3y)^2 - 2(3y)(11z) + (11z)^2 = (3y - 11z)^2$ [Identity II]

(iii) Using $A^2 - B^2 = (A+B)(A-B)$ with $A = 2a - 3b$ and $B = 3a - 2b$:

$A + B = (2a - 3b) + (3a - 2b) = 5a - 5b = 5(a - b)$
$A - B = (2a - 3b) - (3a - 2b) = -a - b = -(a + b)$

Or equivalently: $= -5(a^2 - b^2)$.

Question: Expand:
(i) $(2x + 1)^3$
(ii) $(2a - 3b)^3$

Solution:

(i) Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

(ii) Using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

Question: Factorise $8x^3 + 27y^3 + 36x^2y + 54xy^2$.

Solution:
Rearrange: $8x^3 + 36x^2y + 54xy^2 + 27y^3$

Recognise this as $(a + b)^3$ where $a = 2x$ and $b = 3y$:

Question: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.

Solution:
Using Identity VIII:

Since $x + y + z = 0$:

Application: If $a + b + c = 0$, find $a^3 + b^3 + c^3$.

Using the result above, $a^3 + b^3 + c^3 = 3abc$. This is a guaranteed 1-mark question in exams.

Question: Factorise $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.

Solution:
Rewrite as:

This matches $(a + b + c)^2$ with $a = 2x$, $b = -y$, $c = z$.

Question: Evaluate $(-12)^3 + (7)^3 + (5)^3$ without direct computation.

Solution:
Note that $-12 + 7 + 5 = 0$.

By the identity, when $a + b + c = 0$: $a^3 + b^3 + c^3 = 3abc$.

For dividing by a linear factor $(x - a)$, synthetic division is faster:

Example: Divide $x^3 - 2x^2 - x + 2$ by $(x - 1)$ (i.e., $a = 1$).

Write the coefficients: $1, -2, -1, 2$.

```
1 | 1 -2 -1 2
| 1 -1 -2
| 1 -1 -2 0
```

Process: Bring down 1. Multiply by $a = 1$: $1 \times 1 = 1$. Add: $-2 + 1 = -1$. Multiply: $-1 \times 1 = -1$. Add: $-1 + (-1) = -2$. Multiply: $-2 \times 1 = -2$. Add: $2 + (-2) = 0$.

Quotient coefficients: $1, -1, -2$ → $x^2 - x - 2$. Remainder: $0$.