Chapter 2: Polynomials

Concept used

For any polynomial $$y = p(x)$$, a zero (or root) is a value of $$x$$ for which $$p(x)=0$$. On the cartesian plane this means the point lies on the x-axis (because on the x-axis, $$y=0$$). Therefore:

The number of zeroes of $$p(x)$$ equals the number of points where its graph intersects or just touches the x-axis.

Look at each part of Fig. 2.9 and simply count those contact points.

Thus each graph gives the required count of zeroes as summarised above.

Concept used
Zeroes (or roots) of a polynomial $$p(x)$$ are the values of $$x$$ for which $$p(x)=0$$. On the coordinate plane the condition $$p(x)=0$$ corresponds to the points where the graph of $$y=p(x)$$ meets or touches the x-axis. Therefore:

• Each point of intersection with the x-axis contributes one zero.
• If the graph merely touches the axis (is tangent) and turns back, that point is still counted as one zero.

So, to find the required number we only have to count how many distinct points the given curve has with the x-axis in every sub-figure of Fig. 2.10.

1. Figure (i): the curve lies completely above the x-axis and never meets it.
   Hence, $$0$$ zeroes.
2. Figure (ii): the curve just touches the x-axis at one point and turns back.
   Hence, $$1$$ zero.
3. Figure (iii): the curve cuts the x-axis at two distinct points.
   Hence, $$2$$ zeroes.
4. Figure (iv): the curve meets the x-axis at three distinct points.
   Hence, $$3$$ zeroes.
5. Figure (v): the curve crosses the x-axis at four distinct points.
   Hence, $$4$$ zeroes.
6. Figure (vi): the curve intersects the x-axis at five distinct points.
   Hence, $$5$$ zeroes.

Thus each graph illustrates a different possible number of real zeroes for a polynomial.

The given quadratic polynomial is $$x^{2}+7x+10$$.

Step 1: Identify the coefficients
Compare with the general form $$ax^{2}+bx+c$$. We get
$$a=1,\; b=7,\; c=10$$.

Step 2: Factorise the polynomial

• We need two numbers whose product is $$a\,c = 1 \times 10 = 10$$ and whose sum is $$b = 7$$.
• The numbers $$5$$ and $$2$$ work because $$5+2 = 7$$ and $$5\times 2 = 10$$.

Rewrite the middle term using these numbers:

$$x^{2}+7x+10 = x^{2}+5x+2x+10$$

Group the terms:

$$x^{2}+5x+2x+10 = (x^{2}+5x)+(2x+10)$$

Factor each group:

$$ (x^{2}+5x)+(2x+10) = x(x+5)+2(x+5)$$

Take the common binomial factor $$x+5$$ out:

$$x(x+5)+2(x+5) = (x+5)(x+2)$$

Step 3: Find the zeroes

Set each factor equal to zero:

$$x+5 = 0 \;\;\Rightarrow\;\; x = -5$$
$$x+2 = 0 \;\;\Rightarrow\;\; x = -2$$

Thus, the zeroes are $$\alpha = -5$$ and $$\beta = -2$$.

Step 4: Verify the relationship between zeroes and coefficients

For any quadratic $$ax^{2}+bx+c$$ with zeroes $$\alpha,\beta$$,

Sum of zeroes: $$\alpha+\beta = -\dfrac{b}{a}$$
Product of zeroes: $$\alpha\beta = \dfrac{c}{a}$$

Here, $$a=1,\; b=7,\; c=10$$.

• Sum: $$\alpha+\beta = (-5)+(-2) = -7$$
  Right-hand side: $$-\dfrac{b}{a} = -\dfrac{7}{1} = -7$$
  The two values match.
• Product: $$\alpha\beta = (-5)(-2) = 10$$
  Right-hand side: $$\dfrac{c}{a} = \dfrac{10}{1} = 10$$
  The two values match.

Both relationships are satisfied, so the verification is complete.

The given quadratic polynomial is $$p(x)=x^{2}-3$$.

Step 1: Find the zeroes of $$p(x)$$

Set the polynomial equal to zero:

$$x^{2}-3=0$$

Transpose $$-3$$ to the right side:

$$x^{2}=3$$

Taking the square root of both sides gives two values:

$$x=\pm\sqrt{3}$$

Thus the zeroes are $$\alpha=\sqrt{3}$$ and $$\beta=-\sqrt{3}$$.

Step 2: Write down the coefficients

The standard form of a quadratic is $$ax^{2}+bx+c$$. Comparing with $$x^{2}-3$$ we get:

• $$a=1$$
• $$b=0$$
• $$c=-3$$

Step 3: Verify the relationship between zeroes and coefficients

For any quadratic polynomial:

• Sum of the zeroes: $$\alpha+\beta=-\frac{b}{a}$$
• Product of the zeroes: $$\alpha\beta=\frac{c}{a}$$

Using the calculated zeroes

Sum: $$\alpha+\beta=\sqrt{3}+(-\sqrt{3})=0$$

Product: $$\alpha\beta=\sqrt{3}\times(-\sqrt{3})=-3$$

Using the coefficient relations

Sum: $$-\frac{b}{a}=-\frac{0}{1}=0$$

Product: $$\frac{c}{a}=\frac{-3}{1}=-3$$

The sums match and the products match, therefore the relationship between the zeroes and the coefficients is verified.

Let the quadratic polynomial be monic (leading coefficient equal to $$1$$).
Denote its zeroes by $$\alpha$$ and $$\beta$$.

For any quadratic polynomial $$ax^2+bx+c$$ with $$a\neq0$$, the relations between coefficients and zeroes are

• sum of zeroes: $$\alpha+\beta=-\frac{b}{a}$$,
• product of zeroes: $$\alpha\beta=\frac{c}{a}$$.

When $$a=1$$, we can write the polynomial directly as

$$x^2-(\alpha+\beta)x+\alpha\beta$$.

The problem gives

$$\alpha+\beta=-3,\qquad \alpha\beta=2.$$

Substituting these values, we obtain

$$x^2-(-3)x+2=x^2+3x+2.$$

Hence the required quadratic polynomial is $$x^2+3x+2$$.

Check: $$x^2+3x+2=(x+1)(x+2)$$ has zeroes $$-1$$ and $$-2$$. Their sum $$(-1)+(-2)=-3$$ and product $$(-1)(-2)=2$$, exactly as required.

Step 1 : Write the polynomial
$$p(x)=3x^{3}-5x^{2}-11x-3$$

Step 2 : Verify each proposed zero

(i) For $$x=3$$:
$$p(3)=3(3)^{3}-5(3)^{2}-11(3)-3=3\cdot27-5\cdot9-33-3=81-45-33-3=0$$

(ii) For $$x=-1$$:
$$p(-1)=3(-1)^{3}-5(-1)^{2}-11(-1)-3=-3-5+11-3=0$$

(iii) For $$x=-\dfrac13$$:
$$p\!\left(-\dfrac13\right)=3\left(-\dfrac1{27}\right)-5\left(\dfrac1{9}\right)-11\left(-\dfrac13\right)-3=-\dfrac19-\dfrac59+\dfrac{11}3-3=-\dfrac23+\dfrac{11}3-3=3-3=0$$

Therefore $$3,-1,-\dfrac13$$ are indeed the zeroes of $$p(x)$$.

Step 3 : State the general relationships
For a cubic $$ax^{3}+bx^{2}+cx+d$$ with zeroes $$\alpha ,\beta ,\gamma$$:

• $$\alpha+\beta+\gamma=-\dfrac ba$$
• $$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac ca$$
• $$\alpha\beta\gamma=-\dfrac da$$

Here $$a=3,\;b=-5,\;c=-11,\;d=-3$$.

Step 4 : Compute the left-hand sides with the found zeroes

Sum of zeroes:
$$\alpha+\beta+\gamma=3+(-1)+\left(-\dfrac13\right)=\dfrac53$$

Sum of pairwise products:
$$\alpha\beta+\beta\gamma+\gamma\alpha=3(-1)+(-1)\left(-\dfrac13\right)+\left(-\dfrac13\right)3=-3+\dfrac13-1=-\dfrac{11}3$$

Product of zeroes:
$$\alpha\beta\gamma=3\times(-1)\times\left(-\dfrac13\right)=1$$

Step 5 : Compute the right-hand sides from coefficients

$$-\dfrac ba=-\dfrac{-5}3=\dfrac53$$
$$\dfrac ca=\dfrac{-11}3=-\dfrac{11}3$$
$$-\dfrac da=-\dfrac{-3}3=1$$

Step 6 : Compare
The left-hand and right-hand sides match in every case, so the relationships between the zeroes and the coefficients are verified.