Polynomials

1. What a polynomial is (recap from Class 9)

A polynomial in one variable $x$ is a sum of terms of the form (number) $\times$ (whole-number power of $x$). The degree is the highest power of $x$ that appears. Examples:

• $4x+2$ — degree $1$.
• $2y^{2}-3y+4$ — degree $2$ (in $y$).
• $5x^{3}-4x^{2}+x-\sqrt2$ — degree $3$.
• $7u^{6}-\dfrac{3}{2}u^{4}+4u^{2}+u-8$ — degree $6$.

Not polynomials (powers are negative, fractional, or the variable sits in a denominator): $\dfrac{1}{x-1},\ \sqrt{x}+2,\ \dfrac{1}{x^{2}+2x+3}$.

2. Naming polynomials by degree

• Linear (degree 1): general form $ax+b,\ a\neq0$. E.g. $2x-3,\ \sqrt3x+5,\ y+\sqrt2$.
• Quadratic (degree 2): general form $ax^{2}+bx+c,\ a\neq0$. The word "quadratic" comes from quadrate = square. E.g. $2x^{2}+3x-\dfrac{2}{5},\ \sqrt5v^{2}-\dfrac{2}{3}v$.
• Cubic (degree 3): general form $ax^{3}+bx^{2}+cx+d,\ a\neq0$. E.g. $2-x^{3},\ 3x^{3}-2x^{2}+x-1$.

In every general form, $a,b,c,d$ are real numbers and the leading coefficient $a\neq0$ (otherwise the degree would drop).

3. Value of a polynomial and the idea of a zero

If $p(x)$ is a polynomial and $k$ is a real number, the value of $p(x)$ at $x=k$ is found by replacing $x$ with $k$; we write it $p(k)$. Take $p(x)=x^{2}-3x-4$:

Now test two special inputs:

Because $p(-1)=0$ and $p(4)=0$, we call $-1$ and $4$ the zeroes of $p(x)$.

4. Zero of a linear polynomial

For a linear polynomial $ax+b$, set it equal to $0$: $ax+b=0\Rightarrow x=\dfrac{-b}{a}$. So a linear polynomial has exactly one zero.

Example: zero of $2x+3$ is $k$ where $2k+3=0$, i.e. $k=-\dfrac{3}{2}$. Notice the zero is already linked to the coefficients — this chapter extends that link to quadratics and cubics.

5. Geometrical meaning — linear polynomial

The graph of $y=ax+b$ is a straight line. For $y=2x+3$, plot a couple of points:

The line passes through $(-2,-1)$ and $(2,7)$, and it crosses the x-axis at $\left(-\dfrac{3}{2},0\right)$. That x-coordinate, $-\dfrac{3}{2}$, is exactly the zero of $2x+3$.

General rule: the line $y=ax+b$ meets the x-axis at one point $\left(\dfrac{-b}{a},0\right)$, so a linear polynomial has exactly one zero — the x-coordinate of that crossing point.

6. Geometrical meaning — quadratic polynomial

The graph of $y=ax^{2}+bx+c$ is a curve called a parabola. It opens upwards ($\bigcup$) when $a>0$ and downwards ($\bigcap$) when $a<0$. Take $y=x^{2}-3x-4$ and tabulate:

The parabola cuts the x-axis at $x=-1$ and $x=4$ — exactly the zeroes found in §3. So:

The zeroes of a quadratic are the x-coordinates of the points where the parabola meets the x-axis. Three cases can happen:

• Case (i): the parabola cuts the x-axis at two distinct points $\Rightarrow$ two distinct zeroes.
• Case (ii): it touches the x-axis at exactly one point (vertex on the axis) $\Rightarrow$ two equal (coincident) zeroes, i.e. one zero.
• Case (iii): it stays entirely above or below the x-axis $\Rightarrow$ no real zero.

So a quadratic has at most 2 zeroes.

7. Geometrical meaning — cubic polynomial

For a cubic, the graph can cross the x-axis at up to three points. Consider $y=x^{3}-4x$:

It cuts the x-axis at $-2,\ 0,\ 2$ — these are its three zeroes. Other cubics behave differently: $y=x^{3}$ meets the axis only at $0$ (one zero), and $y=x^{3}-x^{2}=x^{2}(x-1)$ meets it at $0$ and $1$ (two zeroes). A cubic therefore has at most 3 zeroes.

General Remark: a polynomial $p(x)$ of degree $n$ has graph cutting the x-axis at at most $n$ points, so it has at most $n$ zeroes.

8. Relationship between zeroes and coefficients — quadratic

This is the heart of the chapter. Take $p(x)=2x^{2}-8x+6$ and factorise by splitting the middle term (split $-8x$ into two terms whose product is $6\times2x^{2}=12x^{2}$, namely $-6x$ and $-2x$):

So $p(x)=0$ when $x=1$ or $x=3$; the zeroes are $\alpha=1,\ \beta=3$. Observe:

Why it works in general: if $\alpha,\beta$ are the zeroes of $ax^{2}+bx+c$, then $(x-\alpha)$ and $(x-\beta)$ are factors, so $ax^{2}+bx+c=k(x-\alpha)(x-\beta)=k[x^{2}-(\alpha+\beta)x+\alpha\beta]$. Comparing coefficients gives $a=k,\ b=-k(\alpha+\beta),\ c=k\alpha\beta$. Dividing:

9. Building a quadratic from its zeroes

Reverse the relation: if you know the sum $S=\alpha+\beta$ and the product $P=\alpha\beta$, the quadratic is

for any non-zero constant $k$ (usually take $k=1$). There are infinitely many such polynomials (all multiples of one another).

10. Relationship for a cubic (good to know)

For a cubic $ax^{3}+bx^{2}+cx+d$ with zeroes $\alpha,\beta,\gamma$ (gamma):

Take $p(x)=2x^{3}-5x^{2}-14x+8$ with zeroes $4,\ -2,\ \dfrac{1}{2}$. Then sum $=4-2+\dfrac{1}{2}=\dfrac{5}{2}=\dfrac{-(-5)}{2}$; product $=4\times(-2)\times\dfrac{1}{2}=-4=\dfrac{-8}{2}$; and the two-at-a-time sum $=\{4\times(-2)\}+\{(-2)\times\tfrac12\}+\{\tfrac12\times4\}=-8-1+2=-7=\dfrac{-14}{2}$. All three match.

11. NCERT Exercise 2.1 — fully solved

Q1. The graphs of $y=p(x)$ in Fig. 2.10 are given; find the number of zeroes in each case. The number of zeroes = number of points where the graph cuts (or touches) the x-axis.

• (i) The graph does not cut the x-axis at all $\Rightarrow$ 0 zeroes.
• (ii) Cuts the x-axis at 1 point $\Rightarrow$ 1 zero.
• (iii) Cuts the x-axis at 3 points $\Rightarrow$ 3 zeroes.
• (iv) Cuts the x-axis at 2 points $\Rightarrow$ 2 zeroes.
• (v) Cuts the x-axis at 4 points $\Rightarrow$ 4 zeroes.
• (vi) Cuts the x-axis at 3 points $\Rightarrow$ 3 zeroes.

12. NCERT Exercise 2.2 — fully solved

Q1. Find the zeroes and verify the relation between zeroes and coefficients.

(i) $x^{2}-2x-8$. Split $-2x=-4x+2x$: $x^{2}-4x+2x-8=x(x-4)+2(x-4)=(x-4)(x+2)$. Zeroes $4,\ -2$. Sum $=4+(-2)=2=\dfrac{-(-2)}{1}$; product $=4\times(-2)=-8=\dfrac{-8}{1}$. ✓

(ii) $4s^{2}-4s+1$. This is $(2s-1)^{2}$. Zeroes $\dfrac12,\ \dfrac12$ (equal). Sum $=\dfrac12+\dfrac12=1=\dfrac{-(-4)}{4}$; product $=\dfrac12\times\dfrac12=\dfrac14=\dfrac{1}{4}$. ✓

(iii) $6x^{2}-3-7x=6x^{2}-7x-3$. Split $-7x=-9x+2x$: $6x^{2}-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1)$. Zeroes $\dfrac32,\ -\dfrac13$. Sum $=\dfrac32-\dfrac13=\dfrac76=\dfrac{-(-7)}{6}$; product $=\dfrac32\times\left(-\dfrac13\right)=-\dfrac12=\dfrac{-3}{6}$. ✓

(iv) $4u^{2}+8u=4u(u+2)$. Zeroes $0,\ -2$. Sum $=0+(-2)=-2=\dfrac{-8}{4}$; product $=0\times(-2)=0=\dfrac{0}{4}$ (constant term is $0$). ✓

(v) $t^{2}-15=(t-\sqrt{15})(t+\sqrt{15})$. Zeroes $\sqrt{15},\ -\sqrt{15}$. Sum $=0=\dfrac{-0}{1}$; product $=-15=\dfrac{-15}{1}$. ✓

(vi) $3x^{2}-x-4$. Split $-x=-4x+3x$: $3x^{2}-4x+3x-4=x(3x-4)+1(3x-4)=(3x-4)(x+1)$. Zeroes $\dfrac43,\ -1$. Sum $=\dfrac43-1=\dfrac13=\dfrac{-(-1)}{3}$; product $=\dfrac43\times(-1)=-\dfrac43=\dfrac{-4}{3}$. ✓

Q2. Find a quadratic polynomial with the given sum ($S$) and product ($P$) of zeroes. Use $x^{2}-Sx+P$.

• (i) $S=\dfrac14,\ P=-1$: $x^{2}-\dfrac14x-1$, i.e. $4x^{2}-x-4$.
• (ii) $S=\sqrt2,\ P=\dfrac13$: $x^{2}-\sqrt2\,x+\dfrac13$, i.e. $3x^{2}-3\sqrt2\,x+1$.
• (iii) $S=0,\ P=\sqrt5$: $x^{2}+\sqrt5$.
• (iv) $S=1,\ P=1$: $x^{2}-x+1$.
• (v) $S=-\dfrac14,\ P=\dfrac14$: $x^{2}+\dfrac14x+\dfrac14$, i.e. $4x^{2}+x+1$.
• (vi) $S=4,\ P=1$: $x^{2}-4x+1$.

13. Common mistakes to avoid

• Sign of the sum: sum of zeroes is $\dfrac{-b}{a}$, not $\dfrac{b}{a}$ — forgetting the minus sign is the most common slip.
• Mixing up the two formulae: sum $=\dfrac{-b}{a}$, product $=\dfrac{c}{a}$ (not the other way round).
• Treating $\dfrac1x,\ \sqrt x$ as polynomials — they are not (powers must be whole numbers).
• When building a quadratic, writing $x^{2}+Sx+P$ instead of $x^{2}-Sx+P$ — the middle sign is minus the sum.
• Saying a quadratic "always has 2 zeroes" — it can have 2, 1 (equal), or 0 real zeroes.
• Counting zeroes from a graph by where it touches the y-axis — it is the x-axis crossings that matter.

14. Quick revision checklist

• Degree names: 1 = linear, 2 = quadratic, 3 = cubic.
• $k$ is a zero of $p(x)\iff p(k)=0\iff$ graph meets x-axis at $x=k$.
• Degree $n\Rightarrow$ at most $n$ zeroes.
• Quadratic $ax^{2}+bx+c$: $\alpha+\beta=\dfrac{-b}{a},\ \alpha\beta=\dfrac{c}{a}$.
• Build from zeroes: $k[x^{2}-Sx+P]$ where $S$ = sum, $P$ = product.
• Cubic $ax^{3}+bx^{2}+cx+d$: sum $=\dfrac{-b}{a}$, two-at-a-time $=\dfrac{c}{a}$, product $=\dfrac{-d}{a}$.