EXAMPLES
Polynomials • 5 Questions
Question 1
Hint available
Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x). Fig. 2.9
Key Idea
A zero (root) of a polynomial p(x) is a value of x for which p(x)=0. Graphically, this corresponds to the points where the curve y = p(x) intersects the x‑axis. Hence, the number of zeroes equals the number of distinct x‑intercepts of the graph.
Step-by-Step Solution
1. Identify the x‑axis – In each sub‑figure (a), (b), (c) and (d) draw an imaginary horizontal line representing the x‑axis.
2. Count the intersections – Observe how many times the curve meets this line.
- Figure (a): The curve cuts the x‑axis at two distinct points. Therefore, p(x) has 2 zeroes.
- Figure (b): The curve touches the x‑axis at exactly one point (tangent) and does not cross it elsewhere. Hence, p(x) has 1 zero.
- Figure (c): The curve crosses the x‑axis three times, giving three distinct points where p(x)=0. Thus, p(x) has 3 zeroes.
- Figure (d): The curve never meets the x‑axis; it stays entirely above (or below) it. Consequently, p(x) has 0 zeroes.
3. Write the answers – Summarize the counts for each graph.
- (a) 2 zeroes
- (b) 1 zero
- (c) 3 zeroes
- (d) 0 zeroes
2. Count the intersections – Observe how many times the curve meets this line.
- Figure (a): The curve cuts the x‑axis at two distinct points. Therefore, p(x) has 2 zeroes.
- Figure (b): The curve touches the x‑axis at exactly one point (tangent) and does not cross it elsewhere. Hence, p(x) has 1 zero.
- Figure (c): The curve crosses the x‑axis three times, giving three distinct points where p(x)=0. Thus, p(x) has 3 zeroes.
- Figure (d): The curve never meets the x‑axis; it stays entirely above (or below) it. Consequently, p(x) has 0 zeroes.
3. Write the answers – Summarize the counts for each graph.
- (a) 2 zeroes
- (b) 1 zero
- (c) 3 zeroes
- (d) 0 zeroes
Question 2
Hint available
Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Key Idea
For a quadratic polynomial $ax^2+bx+c=0$, if its zeroes are $\alpha$ and $\beta$, then the sum and product of the zeroes are given by the relationships $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$. These are derived from factorising the polynomial as $a(x-\alpha)(x-\beta)$.
Step-by-Step Solution
1. Write the given polynomial: \[f(x)=x^{2}+7x+10\] Here, $a=1$, $b=7$, $c=10$.
2. Find the zeroes by factorisation:
\[x^{2}+7x+10 = (x+5)(x+2)\]
Set each factor to zero:
\[x+5=0 \;\Rightarrow\; x=-5\]
\[x+2=0 \;\Rightarrow\; x=-2\]
Hence the zeroes are $\alpha=-5$ and $\beta=-2$.
3. Verify the sum of zeroes:
\[\alpha+\beta = (-5)+(-2) = -7\]
According to the theorem, \[\alpha+\beta = -\frac{b}{a}= -\frac{7}{1}= -7\]
The calculated sum matches the theoretical value.
4. Verify the product of zeroes:
\[\alpha\beta = (-5)(-2) = 10\]
According to the theorem, \[\alpha\beta = \frac{c}{a}= \frac{10}{1}=10\]
The calculated product also matches the theoretical value.
5. Conclusion: The zeroes $-5$ and $-2$ satisfy both the sum and product relationships with the coefficients of the polynomial.
2. Find the zeroes by factorisation:
\[x^{2}+7x+10 = (x+5)(x+2)\]
Set each factor to zero:
\[x+5=0 \;\Rightarrow\; x=-5\]
\[x+2=0 \;\Rightarrow\; x=-2\]
Hence the zeroes are $\alpha=-5$ and $\beta=-2$.
3. Verify the sum of zeroes:
\[\alpha+\beta = (-5)+(-2) = -7\]
According to the theorem, \[\alpha+\beta = -\frac{b}{a}= -\frac{7}{1}= -7\]
The calculated sum matches the theoretical value.
4. Verify the product of zeroes:
\[\alpha\beta = (-5)(-2) = 10\]
According to the theorem, \[\alpha\beta = \frac{c}{a}= \frac{10}{1}=10\]
The calculated product also matches the theoretical value.
5. Conclusion: The zeroes $-5$ and $-2$ satisfy both the sum and product relationships with the coefficients of the polynomial.
Question 3
Hint available
Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.
Key Idea
For a quadratic polynomial $ax^2+bx+c$, the zeroes (roots) $\alpha$ and $\beta$ satisfy $\alpha+\beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$. The roots can be obtained using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Step-by-Step Solution
1. Identify the coefficients\
The given polynomial is $x^2-3=0$.\
Hence $a=1$, $b=0$, $c=-3$.\
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2. Compute the discriminant\
$$\Delta = b^2-4ac = 0^2-4(1)(-3)=12.$$\
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3. Apply the quadratic formula\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-0 \pm \sqrt{12}}{2\cdot1}=\frac{\pm \sqrt{12}}{2}=\pm \frac{\sqrt{12}}{2}.$$\
Simplify $\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}$, therefore\
$$x = \pm \frac{2\sqrt{3}}{2}=\pm \sqrt{3}.$$\
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4. Zeroes (roots)\
$$\alpha = \sqrt{3}, \qquad \beta = -\sqrt{3}.$$\
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5. Verify the relationships\
- Sum of the zeroes: $\alpha+\beta = \sqrt{3}+(-\sqrt{3}) = 0 = -\dfrac{b}{a}$ (since $b=0$).\
- Product of the zeroes: $\alpha\beta = (\sqrt{3})(-\sqrt{3}) = -3 = \dfrac{c}{a}$ (since $c=-3$).\
Both relationships hold, confirming the theory.
The given polynomial is $x^2-3=0$.\
Hence $a=1$, $b=0$, $c=-3$.\
\
2. Compute the discriminant\
$$\Delta = b^2-4ac = 0^2-4(1)(-3)=12.$$\
\
3. Apply the quadratic formula\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-0 \pm \sqrt{12}}{2\cdot1}=\frac{\pm \sqrt{12}}{2}=\pm \frac{\sqrt{12}}{2}.$$\
Simplify $\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}$, therefore\
$$x = \pm \frac{2\sqrt{3}}{2}=\pm \sqrt{3}.$$\
\
4. Zeroes (roots)\
$$\alpha = \sqrt{3}, \qquad \beta = -\sqrt{3}.$$\
\
5. Verify the relationships\
- Sum of the zeroes: $\alpha+\beta = \sqrt{3}+(-\sqrt{3}) = 0 = -\dfrac{b}{a}$ (since $b=0$).\
- Product of the zeroes: $\alpha\beta = (\sqrt{3})(-\sqrt{3}) = -3 = \dfrac{c}{a}$ (since $c=-3$).\
Both relationships hold, confirming the theory.
Question 4
Hint available
Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.
Key Idea
Use Vieta’s formulas for a quadratic polynomial $ax^2+bx+c=0$: sum of zeros $\alpha+\beta = -\frac{b}{a}$ and product of zeros $\alpha\beta = \frac{c}{a}$. Choose $a=1$ for simplicity and equate the given sum and product to find $b$ and $c$.
Step-by-Step Solution
1. Let the required quadratic polynomial be \(ax^2+bx+c=0\).
2. For its zeros \(\alpha\) and \(\beta\), Vieta’s relations give:
$$\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.$$
3. The problem states \(\alpha+\beta = -3\) and \(\alpha\beta = 2\).
4. Choose the simplest leading coefficient \(a=1\) (any non‑zero constant multiple would also work). Then the relations become:
$$-b = -3 \;\Rightarrow\; b = 3,$$
$$c = 2.$$
5. Substitute \(a=1,\; b=3,\; c=2\) into the general form:
$$x^2 + 3x + 2 = 0.$$
6. Hence the required quadratic polynomial (up to a non‑zero constant factor) is \(x^2 + 3x + 2\).
2. For its zeros \(\alpha\) and \(\beta\), Vieta’s relations give:
$$\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.$$
3. The problem states \(\alpha+\beta = -3\) and \(\alpha\beta = 2\).
4. Choose the simplest leading coefficient \(a=1\) (any non‑zero constant multiple would also work). Then the relations become:
$$-b = -3 \;\Rightarrow\; b = 3,$$
$$c = 2.$$
5. Substitute \(a=1,\; b=3,\; c=2\) into the general form:
$$x^2 + 3x + 2 = 0.$$
6. Hence the required quadratic polynomial (up to a non‑zero constant factor) is \(x^2 + 3x + 2\).
Question 5
Hint available
* : Verify that 3, –1, are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Key Idea
Use the Factor Theorem to check each given number as a zero of the polynomial. Then express the cubic as a·(x‑α)(x‑β)(x‑γ) and apply Vieta’s relations: α+β+γ = –b/a, αβ+βγ+γα = c/a, αβγ = –d/a.
Step-by-Step Solution
1. Check the given numbers as zeroes\
• For \(x = 3\):\
\[p(3)=3(3)^3-5(3)^2-11(3)-3=81-45-33-3=0\]\
Hence \(x=3\) is a zero.\
• For \(x = -1\):\
\[p(-1)=3(-1)^3-5(-1)^2-11(-1)-3=-3-5+11-3=0\]\
Hence \(x=-1\) is a zero.\
• To find the third zero, divide \(p(x)\) by \((x-3)(x+1)\).\
First divide by \(x-3\) (synthetic division):\
\[\begin{array}{c|cccc}
3 & 3 & -5 & -11 & -3 \\
& & 9 & 12 & 3 \\
\hline
& 3 & 4 & 1 & 0 \\
\end{array}\]
Quotient = \(3x^2+4x+1\).\
Factor the quadratic:\
\[3x^2+4x+1 = (3x+1)(x+1)\]\
Hence \(p(x)= (x-3)(x+1)(3x+1)\).\
The remaining zero is obtained from \(3x+1=0\) ⇒ \(x = -\frac{1}{3}\).\
Therefore the three zeroes are \(\alpha=3,\; \beta=-1,\; \gamma=-\frac13\).\
2. Verify Vieta’s relations\
The cubic is of the form \(ax^3+bx^2+cx+d\) with \(a=3,\; b=-5,\; c=-11,\; d=-3\).\
- Sum of zeroes:\
\[\alpha+\beta+\gamma = 3+(-1)+\left(-\frac13\right)=\frac{5}{3}\]\
\[-\frac{b}{a}= -\frac{-5}{3}=\frac{5}{3}\]\
Both are equal, so the first relation holds.\
- Sum of products of zeroes taken two at a time:\
\[\alpha\beta+\beta\gamma+\gamma\alpha = (3)(-1)+(-1)\left(-\frac13\right)+(3)\left(-\frac13\right) = -3+\frac13-1 = -\frac{11}{3}\]\
\[\frac{c}{a}= \frac{-11}{3}= -\frac{11}{3}\]\
Hence the second relation is satisfied.\
- Product of the three zeroes:\
\[\alpha\beta\gamma = 3\times(-1)\times\left(-\frac13\right)=1\]\
\[-\frac{d}{a}= -\frac{-3}{3}=1\]\
The third relation also holds.\
Thus, the numbers \(3, -1, -\frac13\) are indeed the zeroes of \(p(x)\) and they satisfy the standard relationships between zeroes and coefficients for a cubic polynomial.
• For \(x = 3\):\
\[p(3)=3(3)^3-5(3)^2-11(3)-3=81-45-33-3=0\]\
Hence \(x=3\) is a zero.\
• For \(x = -1\):\
\[p(-1)=3(-1)^3-5(-1)^2-11(-1)-3=-3-5+11-3=0\]\
Hence \(x=-1\) is a zero.\
• To find the third zero, divide \(p(x)\) by \((x-3)(x+1)\).\
First divide by \(x-3\) (synthetic division):\
\[\begin{array}{c|cccc}
3 & 3 & -5 & -11 & -3 \\
& & 9 & 12 & 3 \\
\hline
& 3 & 4 & 1 & 0 \\
\end{array}\]
Quotient = \(3x^2+4x+1\).\
Factor the quadratic:\
\[3x^2+4x+1 = (3x+1)(x+1)\]\
Hence \(p(x)= (x-3)(x+1)(3x+1)\).\
The remaining zero is obtained from \(3x+1=0\) ⇒ \(x = -\frac{1}{3}\).\
Therefore the three zeroes are \(\alpha=3,\; \beta=-1,\; \gamma=-\frac13\).\
2. Verify Vieta’s relations\
The cubic is of the form \(ax^3+bx^2+cx+d\) with \(a=3,\; b=-5,\; c=-11,\; d=-3\).\
- Sum of zeroes:\
\[\alpha+\beta+\gamma = 3+(-1)+\left(-\frac13\right)=\frac{5}{3}\]\
\[-\frac{b}{a}= -\frac{-5}{3}=\frac{5}{3}\]\
Both are equal, so the first relation holds.\
- Sum of products of zeroes taken two at a time:\
\[\alpha\beta+\beta\gamma+\gamma\alpha = (3)(-1)+(-1)\left(-\frac13\right)+(3)\left(-\frac13\right) = -3+\frac13-1 = -\frac{11}{3}\]\
\[\frac{c}{a}= \frac{-11}{3}= -\frac{11}{3}\]\
Hence the second relation is satisfied.\
- Product of the three zeroes:\
\[\alpha\beta\gamma = 3\times(-1)\times\left(-\frac13\right)=1\]\
\[-\frac{d}{a}= -\frac{-3}{3}=1\]\
The third relation also holds.\
Thus, the numbers \(3, -1, -\frac13\) are indeed the zeroes of \(p(x)\) and they satisfy the standard relationships between zeroes and coefficients for a cubic polynomial.