Polynomials — NCERT Solutions

The number of zeroes of $p(x)$ equals the number of times the graph of $y = p(x)$ intersects (or touches) the $x$-axis.

(i) The graph does not intersect the $x$-axis at all.
$$\text{Number of zeroes} = 0$$

(ii) The graph intersects the $x$-axis at exactly one point.
$$\text{Number of zeroes} = 1$$

(iii) The graph intersects the $x$-axis at exactly three points.
$$\text{Number of zeroes} = 3$$

(iv) The graph intersects the $x$-axis at exactly two points.
$$\text{Number of zeroes} = 2$$

(v) The graph intersects the $x$-axis at exactly four points.
$$\text{Number of zeroes} = 4$$

(vi) The graph intersects the $x$-axis at exactly three points.
$$\text{Number of zeroes} = 3$$

Concept: The zeroes of a quadratic polynomial $ax^2+bx+c$ are found by factorisation (or formula). If $\alpha$ and $\beta$ are the zeroes, then:
$$\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

(i) $p(x) = x^2 - 2x - 8$

Splitting the middle term:
$$x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x-4)+2(x-4) = (x-4)(x+2)$$

Zeroes: $x - 4 = 0 \Rightarrow x = 4$ and $x + 2 = 0 \Rightarrow x = -2$

So $\alpha = 4,\ \beta = -2$.

Verification: Here $a=1,\ b=-2,\ c=-8$.
$$\alpha+\beta = 4+(-2) = 2 = -\frac{(-2)}{1} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = 4\times(-2) = -8 = \frac{-8}{1} = \frac{c}{a} \checkmark$$

(ii) $p(s) = 4s^2 - 4s + 1$

Splitting the middle term:
$$4s^2 - 4s + 1 = 4s^2 - 2s - 2s + 1 = 2s(2s-1)-1(2s-1) = (2s-1)(2s-1)$$

Zeroes: $2s - 1 = 0 \Rightarrow s = \dfrac{1}{2}$ (repeated)

So $\alpha = \beta = \dfrac{1}{2}$.

Verification: Here $a=4,\ b=-4,\ c=1$.
$$\alpha+\beta = \frac{1}{2}+\frac{1}{2} = 1 = -\frac{(-4)}{4} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{c}{a} \checkmark$$

(iii) $p(x) = 6x^2 - 7x - 3$

(Rewriting: $6x^2 - 3 - 7x = 6x^2 - 7x - 3$)

Splitting the middle term (product $= 6\times(-3)=-18$; factors $-9$ and $+2$):
$$6x^2 - 9x + 2x - 3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$$

Zeroes: $3x+1=0 \Rightarrow x = -\dfrac{1}{3}$ and $2x-3=0 \Rightarrow x = \dfrac{3}{2}$

So $\alpha = -\dfrac{1}{3},\ \beta = \dfrac{3}{2}$.

Verification: Here $a=6,\ b=-7,\ c=-3$.
$$\alpha+\beta = -\frac{1}{3}+\frac{3}{2} = \frac{-2+9}{6} = \frac{7}{6} = -\frac{(-7)}{6} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \left(-\frac{1}{3}\right)\times\frac{3}{2} = -\frac{1}{2} = \frac{-3}{6} = \frac{c}{a} \checkmark$$

(iv) $p(u) = 4u^2 + 8u$

Factorising:
$$4u^2 + 8u = 4u(u + 2)$$

Zeroes: $4u = 0 \Rightarrow u = 0$ and $u+2=0 \Rightarrow u = -2$

So $\alpha = 0,\ \beta = -2$.

Verification: Here $a=4,\ b=8,\ c=0$.
$$\alpha+\beta = 0+(-2) = -2 = -\frac{8}{4} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = 0\times(-2) = 0 = \frac{0}{4} = \frac{c}{a} \checkmark$$

(v) $p(t) = t^2 - 15$

Factorising:
$$t^2 - 15 = \left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)$$

Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$

So $\alpha = \sqrt{15},\ \beta = -\sqrt{15}$.

Verification: Here $a=1,\ b=0,\ c=-15$.
$$\alpha+\beta = \sqrt{15}+(-\sqrt{15}) = 0 = -\frac{0}{1} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \sqrt{15}\times(-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a} \checkmark$$

(vi) $p(x) = 3x^2 - x - 4$

Splitting the middle term (product $= 3\times(-4)=-12$; factors $-4$ and $+3$):
$$3x^2 - 4x + 3x - 4 = x(3x-4)+1(3x-4) = (x+1)(3x-4)$$

Zeroes: $x+1=0 \Rightarrow x=-1$ and $3x-4=0 \Rightarrow x=\dfrac{4}{3}$

So $\alpha = -1,\ \beta = \dfrac{4}{3}$.

Verification: Here $a=3,\ b=-1,\ c=-4$.
$$\alpha+\beta = -1+\frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3} = -\frac{(-1)}{3} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = (-1)\times\frac{4}{3} = -\frac{4}{3} = \frac{-4}{3} = \frac{c}{a} \checkmark$$