Polynomials — NCERT Solutions, AI Tutor & Practice

The zeros of a polynomial are precisely the x-coordinates of the points where the graph of $y = p(x)$ intersects the x-axis.

Here the graph cuts the x-axis at two distinct points.

Therefore the polynomial has $\textbf{2 zeros}$.

When a graph touches the x-axis at a point (rather than crossing it), the graph still meets the x-axis at that single point.

So there is exactly 1 point of intersection with the x-axis.

Hence $p(x)$ has $\textbf{1 zero}$ (this is a repeated zero, but geometrically only one distinct point of contact).

A real zero occurs where $y = p(x) = 0$, i.e. where the graph meets the x-axis.

Since the graph never meets the x-axis, there is no value of $x$ for which $p(x) = 0$.

Therefore the polynomial has $\textbf{0 (no) real zeros}$.

The graph of a linear polynomial $ax + b$ (with $a \neq 0$) is a straight line.

A non-horizontal straight line meets the x-axis at exactly one point.

• Degree of the polynomial $= \textbf{1}$ (linear).
• Number of zeros $= \textbf{1}$.

Factorise: $x^2 - 7x + 10 = x^2 - 5x - 2x + 10 = x(x-5) - 2(x-5) = (x-5)(x-2)$.

Zeros: $x = 5$ and $x = 2$.

Verification (for $ax^2 + bx + c$, here $a=1, b=-7, c=10$):

• Sum of zeros $= 5 + 2 = 7$ and $-\dfrac{b}{a} = -\dfrac{-7}{1} = 7$. ✓
• Product of zeros $= 5 \times 2 = 10$ and $\dfrac{c}{a} = \dfrac{10}{1} = 10$. ✓

The relationships hold.

Here $a=6, b=-7, c=-3$, product $ac = -18$. We need two numbers with product $-18$ and sum $-7$: these are $-9$ and $2$.

$6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (2x-3)(3x+1)$.

Zeros: $2x - 3 = 0 \Rightarrow x = \dfrac{3}{2}$; $3x + 1 = 0 \Rightarrow x = -\dfrac{1}{3}$.

Verification:

• Sum $= \dfrac{3}{2} - \dfrac{1}{3} = \dfrac{9-2}{6} = \dfrac{7}{6}$ and $-\dfrac{b}{a} = -\dfrac{-7}{6} = \dfrac{7}{6}$. ✓
• Product $= \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = -\dfrac{1}{2}$ and $\dfrac{c}{a} = \dfrac{-3}{6} = -\dfrac{1}{2}$. ✓

A quadratic polynomial with sum of zeros $S$ and product of zeros $P$ can be written as: $k\left[x^2 - Sx + P\right]$

Here $S = -3$ and $P = 2$: $x^2 - (-3)x + 2 = x^2 + 3x + 2.$

Taking $k = 1$, the required polynomial is $\boxed{x^2 + 3x + 2}$.

(Any non-zero multiple, e.g. $2x^2 + 6x + 4$, is also valid.)

Let the zeros be $\alpha = \dfrac{1}{4}$ and $\beta = -1$.

• Sum $S = \dfrac{1}{4} + (-1) = -\dfrac{3}{4}$.
• Product $P = \dfrac{1}{4} \times (-1) = -\dfrac{1}{4}$.

Polynomial $= x^2 - Sx + P = x^2 + \dfrac{3}{4}x - \dfrac{1}{4}$.

Multiplying by $4$ to clear fractions: $\boxed{4x^2 + 3x - 1}.$

For $x^2 - 5x + 6$ ($a=1, b=-5, c=6$):

• $\alpha + \beta = -\dfrac{b}{a} = 5$
• $\alpha\beta = \dfrac{c}{a} = 6$

Now combine the fractions: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{5}{6}.$

So the value is $\dfrac{5}{6}$.

Since $2$ is a zero, $p(2) = 0$: $2(2)^2 + k(2) + 6 = 0 \Rightarrow 8 + 2k + 6 = 0 \Rightarrow 2k = -14 \Rightarrow k = -7.$

So the polynomial is $2x^2 - 7x + 6$.

Other zero: Product of zeros $= \dfrac{c}{a} = \dfrac{6}{2} = 3$.

If one zero is $2$ and the other is $\beta$, then $2\beta = 3 \Rightarrow \beta = \dfrac{3}{2}$.

Therefore $k = -7$ and the other zero is $\dfrac{3}{2}$.

For $x^2 - 6x + 8$:

• $\alpha + \beta = -\dfrac{-6}{1} = 6$
• $\alpha\beta = \dfrac{8}{1} = 8$

Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $\alpha^2 + \beta^2 = 6^2 - 2(8) = 36 - 16 = 20.$

So $\alpha^2 + \beta^2 = 20$.