Polynomials — NCERT Solutions

Concept: A polynomial in one variable is an expression of the form $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0$ where all exponents of the variable are whole numbers (non-negative integers).

(i) $4x^{2} - 3x + 7$
All exponents of $x$ are whole numbers (2, 1, 0). It is a polynomial in one variable $x$.

(ii) $y^{2} + \sqrt{2}$
All exponents of $y$ are whole numbers (2, 0). It is a polynomial in one variable $y$.

(iii) $3\sqrt{t} + t\sqrt{2}$
$3\sqrt{t} = 3t^{1/2}$. The exponent $\frac{1}{2}$ is not a whole number. Hence it is not a polynomial.

(iv) $y + \frac{2}{y}$
$\frac{2}{y} = 2y^{-1}$. The exponent $-1$ is not a whole number. Hence it is not a polynomial.

(v) $x^{10} + y^3 + t^{50}$
This expression contains three variables $x$, $y$, and $t$. Hence it is not a polynomial in one variable (it is a polynomial in three variables).

Concept: The coefficient of $x^2$ is the number multiplied with $x^2$ in the expression.

(i) $2 + x^{2} + x$
The term containing $x^2$ is $1 \cdot x^2$.
Coefficient of $x^2$ = $\mathbf{1}$

(ii) $2 - x^{2} + x^{3}$
The term containing $x^2$ is $-1 \cdot x^2$.
Coefficient of $x^2$ = $\mathbf{-1}$

(iii) $\frac{\pi}{2} x^2 + x$
The term containing $x^2$ is $\frac{\pi}{2} x^2$.
Coefficient of $x^2$ = $\dfrac{\boldsymbol{\pi}}{\mathbf{2}}$

(iv) $\sqrt{2} x - 1$
There is no $x^2$ term in this expression.
Coefficient of $x^2$ = $\mathbf{0}$

Concept:

• A binomial has exactly two terms.
• A monomial has exactly one term.
• The degree is the highest power of the variable.

Binomial of degree 35:
$$x^{35} + 1$$
This has two terms and the highest power is 35.

Monomial of degree 100:
$$x^{100}$$
This has one term and the highest power is 100.

Concept: The degree of a polynomial is the highest power of the variable in the polynomial.

(i) $5x^{3} + 4x^{2} + 7x$
Highest power of $x$ is 3.
Degree = $\mathbf{3}$

(ii) $4 - y^{2}$
Highest power of $y$ is 2.
Degree = $\mathbf{2}$

(iii) $5t - \sqrt{7}$
Highest power of $t$ is 1.
Degree = $\mathbf{1}$

(iv) $3$
$3 = 3 \cdot x^0$. This is a non-zero constant polynomial.
Degree = $\mathbf{0}$

Concept:

• Linear polynomial: degree 1
• Quadratic polynomial: degree 2
• Cubic polynomial: degree 3

(i) $x^{2} + x$ — Highest degree = 2 → Quadratic polynomial

(ii) $x - x^3$ — Highest degree = 3 → Cubic polynomial

(iii) $y + y^2 + 4$ — Highest degree = 2 → Quadratic polynomial

(iv) $1 + x$ — Highest degree = 1 → Linear polynomial

(v) $3t$ — Highest degree = 1 → Linear polynomial

(vi) $r^2$ — Highest degree = 2 → Quadratic polynomial

(vii) $7x^3$ — Highest degree = 3 → Cubic polynomial

Let $p(x) = 5x - 4x^2 + 3$

(i) At $x = 0$:
$$p(0) = 5(0) - 4(0)^2 + 3 = 0 - 0 + 3 = \mathbf{3}$$

(ii) At $x = -1$:
$$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = \mathbf{-6}$$

(iii) At $x = 2$:
$$p(2) = 5(2) - 4(2)^2 + 3 = 10 - 4(4) + 3 = 10 - 16 + 3 = \mathbf{-3}$$

(i) $p(y) = y^2 - y + 1$
$$p(0) = 0 - 0 + 1 = \mathbf{1}$$
$$p(1) = 1 - 1 + 1 = \mathbf{1}$$
$$p(2) = 4 - 2 + 1 = \mathbf{3}$$

(ii) $p(t) = 2 + t + 2t^2 - t^3$
$$p(0) = 2 + 0 + 0 - 0 = \mathbf{2}$$
$$p(1) = 2 + 1 + 2(1) - 1 = 2 + 1 + 2 - 1 = \mathbf{4}$$
$$p(2) = 2 + 2 + 2(4) - 8 = 2 + 2 + 8 - 8 = \mathbf{4}$$

(iii) $p(x) = x^3$
$$p(0) = 0^3 = \mathbf{0}$$
$$p(1) = 1^3 = \mathbf{1}$$
$$p(2) = 2^3 = \mathbf{8}$$

(iv) $p(x) = (x-1)(x+1)$
$$p(0) = (0-1)(0+1) = (-1)(1) = \mathbf{-1}$$
$$p(1) = (1-1)(1+1) = (0)(2) = \mathbf{0}$$
$$p(2) = (2-1)(2+1) = (1)(3) = \mathbf{3}$$

Concept: $x = a$ is a zero of $p(x)$ if and only if $p(a) = 0$.

(i) $p(x) = 3x + 1$ at $x = -\frac{1}{3}$:
$$p\!\left(-\tfrac{1}{3}\right) = 3\left(-\tfrac{1}{3}\right) + 1 = -1 + 1 = 0$$
Yes, $x = -\dfrac{1}{3}$ is a zero of $p(x)$.

(ii) $p(x) = 5x - \pi$ at $x = \frac{4}{5}$:
$$p\!\left(\tfrac{4}{5}\right) = 5\left(\tfrac{4}{5}\right) - \pi = 4 - \pi \neq 0$$
No, $x = \dfrac{4}{5}$ is not a zero of $p(x)$.

(iii) $p(x) = x^2 - 1$ at $x = 1$ and $x = -1$:
$$p(1) = 1^2 - 1 = 0$$
$$p(-1) = (-1)^2 - 1 = 1 - 1 = 0$$
Yes, both $x = 1$ and $x = -1$ are zeroes of $p(x)$.

(iv) $p(x) = (x+1)(x-2)$ at $x = -1$ and $x = 2$:
$$p(-1) = (-1+1)(-1-2) = (0)(-3) = 0$$
$$p(2) = (2+1)(2-2) = (3)(0) = 0$$
Yes, both $x = -1$ and $x = 2$ are zeroes of $p(x)$.

(v) $p(x) = x^2$ at $x = 0$:
$$p(0) = 0^2 = 0$$
Yes, $x = 0$ is a zero of $p(x)$.

(vi) $p(x) = lx + m$ at $x = -\frac{m}{l}$:
$$p\!\left(-\tfrac{m}{l}\right) = l\left(-\tfrac{m}{l}\right) + m = -m + m = 0$$
Yes, $x = -\dfrac{m}{l}$ is a zero of $p(x)$.

(vii) $p(x) = 3x^2 - 1$ at $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{2}{\sqrt{3}}$:
$$p\!\left(-\tfrac{1}{\sqrt{3}}\right) = 3\left(\tfrac{1}{3}\right) - 1 = 1 - 1 = 0$$
So $x = -\dfrac{1}{\sqrt{3}}$ is a zero.
$$p\!\left(\tfrac{2}{\sqrt{3}}\right) = 3\left(\tfrac{4}{3}\right) - 1 = 4 - 1 = 3 \neq 0$$
So $x = \dfrac{2}{\sqrt{3}}$ is not a zero of $p(x)$.

(viii) $p(x) = 2x + 1$ at $x = \frac{1}{2}$:
$$p\!\left(\tfrac{1}{2}\right) = 2\left(\tfrac{1}{2}\right) + 1 = 1 + 1 = 2 \neq 0$$
No, $x = \dfrac{1}{2}$ is not a zero of $p(x)$.

Concept: To find the zero, set $p(x) = 0$ and solve for $x$.

(i) $p(x) = x + 5$:
$$x + 5 = 0 \implies x = -5$$
Zero is $\mathbf{-5}$.

(ii) $p(x) = x - 5$:
$$x - 5 = 0 \implies x = 5$$
Zero is $\mathbf{5}$.

(iii) $p(x) = 2x + 5$:
$$2x + 5 = 0 \implies 2x = -5 \implies x = -\dfrac{5}{2}$$
Zero is $\mathbf{-\dfrac{5}{2}}$.

(iv) $p(x) = 3x - 2$:
$$3x - 2 = 0 \implies 3x = 2 \implies x = \dfrac{2}{3}$$
Zero is $\mathbf{\dfrac{2}{3}}$.

(v) $p(x) = 3x$:
$$3x = 0 \implies x = 0$$
Zero is $\mathbf{0}$.

(vi) $p(x) = ax,\ a \neq 0$:
$$ax = 0 \implies x = 0$$
Zero is $\mathbf{0}$.

(vii) $p(x) = cx + d,\ c \neq 0$:
$$cx + d = 0 \implies cx = -d \implies x = -\dfrac{d}{c}$$
Zero is $\mathbf{-\dfrac{d}{c}}$.

Concept (Factor Theorem): $(x+1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.

(i) $p(x) = x^3 + x^2 + x + 1$:
$$p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0$$
Since $p(-1) = 0$, $(x+1)$ is a factor.

(ii) $p(x) = x^4 + x^3 + x^2 + x + 1$:
$$p(-1) = 1 - 1 + 1 - 1 + 1 = 1 \neq 0$$
Since $p(-1) \neq 0$, $(x+1)$ is not a factor.

(iii) $p(x) = x^4 + 3x^3 + 3x^2 + x + 1$:
$$p(-1) = 1 + 3(-1) + 3(1) + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1 \neq 0$$
Since $p(-1) \neq 0$, $(x+1)$ is not a factor.

(iv) $p(x) = x^3 - x^2 - (2+\sqrt{2})x + \sqrt{2}$:
$$p(-1) = (-1)^3 - (-1)^2 - (2+\sqrt{2})(-1) + \sqrt{2}$$
$$= -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = 0 + 2\sqrt{2} = 2\sqrt{2} \neq 0$$
Since $p(-1) \neq 0$, $(x+1)$ is not a factor.

Concept: $g(x) = x - a$ is a factor of $p(x)$ iff $p(a) = 0$.

(i) $g(x) = x + 1$, zero is $x = -1$:
$$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0$$
Since $p(-1) = 0$, $g(x)$ is a factor of $p(x)$.

(ii) $g(x) = x + 2$, zero is $x = -2$:
$$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 \neq 0$$
Since $p(-2) \neq 0$, $g(x)$ is not a factor of $p(x)$.

(iii) $g(x) = x - 3$, zero is $x = 3$:
$$p(3) = 3^3 - 4(3)^2 + 3 + 6 = 27 - 36 + 3 + 6 = 0$$
Since $p(3) = 0$, $g(x)$ is a factor of $p(x)$.

Concept: If $(x-1)$ is a factor of $p(x)$, then by Factor Theorem, $p(1) = 0$.

(i) $p(x) = x^2 + x + k$:
$$p(1) = 1 + 1 + k = 0 \implies 2 + k = 0 \implies \mathbf{k = -2}$$

(ii) $p(x) = 2x^2 + kx + \sqrt{2}$:
$$p(1) = 2 + k + \sqrt{2} = 0 \implies k = -2 - \sqrt{2} \implies \mathbf{k = -(2 + \sqrt{2})}$$

(iii) $p(x) = kx^2 - \sqrt{2}x + 1$:
$$p(1) = k - \sqrt{2} + 1 = 0 \implies k = \sqrt{2} - 1 \implies \mathbf{k = \sqrt{2} - 1}$$

(iv) $p(x) = kx^2 - 3x + k$:
$$p(1) = k - 3 + k = 0 \implies 2k = 3 \implies \mathbf{k = \dfrac{3}{2}}$$

Method: Splitting the middle term.

(i) $12x^2 - 7x + 1$:
We need two numbers whose product = $12 \times 1 = 12$ and sum = $-7$.
Numbers: $-3$ and $-4$.
$$12x^2 - 7x + 1 = 12x^2 - 4x - 3x + 1$$
$$= 4x(3x - 1) - 1(3x - 1)$$
$$= \mathbf{(4x - 1)(3x - 1)}$$

(ii) $2x^2 + 7x + 3$:
Product = $2 \times 3 = 6$, sum = $7$. Numbers: $6$ and $1$.
$$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$$
$$= 2x(x + 3) + 1(x + 3)$$
$$= \mathbf{(2x + 1)(x + 3)}$$

(iii) $6x^2 + 5x - 6$:
Product = $6 \times (-6) = -36$, sum = $5$. Numbers: $9$ and $-4$.
$$6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6$$
$$= 3x(2x + 3) - 2(2x + 3)$$
$$= \mathbf{(3x - 2)(2x + 3)}$$

(iv) $3x^2 - x - 4$:
Product = $3 \times (-4) = -12$, sum = $-1$. Numbers: $-4$ and $3$.
$$3x^2 - x - 4 = 3x^2 - 4x + 3x - 4$$
$$= x(3x - 4) + 1(3x - 4)$$
$$= \mathbf{(x + 1)(3x - 4)}$$

Method: Factor Theorem — find a zero by trial, then divide/group.

(i) $p(x) = x^3 - 2x^2 - x + 2$:
$p(1) = 1 - 2 - 1 + 2 = 0$, so $(x-1)$ is a factor.
$$x^3 - 2x^2 - x + 2 = x^2(x-1) - (x-1)(x+2)$$
Let us group:
$$= x^2(x - 1) - (x - 1) - x(x-1)$$
Better: divide $p(x)$ by $(x-1)$:
$$x^3 - 2x^2 - x + 2 = (x-1)(x^2 - x - 2)$$
Now factorise $x^2 - x - 2 = (x-2)(x+1)$.
$$\boxed{(x-1)(x-2)(x+1)}$$

(ii) $p(x) = x^3 - 3x^2 - 9x - 5$:
$p(5) = 125 - 75 - 45 - 5 = 0$, so $(x-5)$ is a factor.
Divide: $x^3 - 3x^2 - 9x - 5 = (x-5)(x^2 + 2x + 1) = (x-5)(x+1)^2$
$$\boxed{(x-5)(x+1)^2}$$

(iii) $p(x) = x^3 + 13x^2 + 32x + 20$:
$p(-1) = -1 + 13 - 32 + 20 = 0$, so $(x+1)$ is a factor.
Divide: $x^3 + 13x^2 + 32x + 20 = (x+1)(x^2 + 12x + 20)$
Factorise $x^2 + 12x + 20 = (x+2)(x+10)$.
$$\boxed{(x+1)(x+2)(x+10)}$$

(iv) $p(y) = 2y^3 + y^2 - 2y - 1$:
$p(1) = 2 + 1 - 2 - 1 = 0$, so $(y-1)$ is a factor.
Group: $2y^3 + y^2 - 2y - 1 = y^2(2y+1) - 1(2y+1) = (2y+1)(y^2-1) = (2y+1)(y-1)(y+1)$
$$\boxed{(2y+1)(y-1)(y+1)}$$

Identity used: $(x+a)(x+b) = x^2 + (a+b)x + ab$ and $(x+y)(x-y) = x^2 - y^2$, $(x-y)^2 = x^2 - 2xy + y^2$.

(i) $(x+4)(x+10)$: Using Identity IV with $a=4, b=10$:
$$= x^2 + (4+10)x + (4)(10) = \mathbf{x^2 + 14x + 40}$$

(ii) $(x+8)(x-10)$: Using Identity IV with $a=8, b=-10$:
$$= x^2 + (8-10)x + (8)(-10) = \mathbf{x^2 - 2x - 80}$$

(iii) $(3x+4)(3x-5)$: Using Identity IV with $x \to 3x$, $a=4$, $b=-5$:
$$= (3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 - 3x - 20 = \mathbf{9x^2 - 3x - 20}$$

(iv) $\left(y^2 - \dfrac{3}{2}\right)^2$: Using Identity II $(x-y)^2 = x^2 - 2xy + y^2$ with $x = y^2$, $y = \dfrac{3}{2}$:
$$= y^4 - 2 \cdot y^2 \cdot \frac{3}{2} + \frac{9}{4} = \mathbf{y^4 - 3y^2 + \dfrac{9}{4}}$$

(v) $(3-2x)(3+2x)$: Using Identity III $(a-b)(a+b) = a^2 - b^2$ with $a=3$, $b=2x$:
$$= 9 - 4x^2 = \mathbf{9 - 4x^2}$$