Explanation
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.
Consider first a linear polynomial ax + b, a ≠ 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).
X |
-2 |
2 |
Y = 2x + 3 |
-1 |
7 |
The graph of y = 2x + 3 intersects the x-axis mid-way between x = –1 and x = – 2,
that is, at the point (-3/2, 0). You also know that the zero of 2x + 3 is -3/2. Thus, the zero of the polynomial 2x + 3 is the x – coordinate of the point point where the graph of y = 2x + 3 intersects the x-axis.
In general, for a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a straight line which intersects the x-axis at exactly one point, namely, (-b/a, 0).
Therefore, the linear polynomial ax + b, a ≠ 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph* of y = x2 – 3x – 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x.
X |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
5 |
Y = x^{2} |
6 |
0 |
-4 |
-6 |
-6 |
-4 |
0 |
6 |
In fact, for any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open
downwards like depending on whether a > 0 or a < 0.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 –3x – 4 intersects the x-axis. Thus, the zeroes of the quadratic polynomial x 2 – 3x – 4 are x-coordinates of the points where the graph of y = x 2 – 3x – 4 intersects the x -axis.
Case (i) : Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax^{2}
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point.
So, the quadratic polynomial ax^{2}
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be? Let us find out. Consider the cubic polynomial x^{3}
X |
-2 |
-1 |
0 |
1 |
2 |
Y = x^{3} |
0 |
3 |
0 |
-3 |
0 |
Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x^{3}
We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x. Observe that –2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x^{3}
Let us take a few more examples. Consider the cubic polynomials x^{3}
Note that 0 is the only zero of the polynomial x^{3}
Example 1 : Look at the graphs. 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).
(i)The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii)The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii)The number of zeroes is 3.
(iv)The number of zeroes is 1.
(v)The number of zeroes is 1.
(vi)The number of zeroes is 4.
You have already seen that zero of a linear polynomial ax + b is \frac{b}{a}
2x^{2}
= (2x – 2)(x – 3) = 2(x – 1)(x – 3)
Example 2 : Find the zeroes of the quadratic polynomial x 2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Solution : We have
x^{2}
So, the value of x^{2}
Sum of zeroes = -2 + (-5) = -(7) = \frac{-(7)}{1} = \frac{-coefficient \:\: of \:\: x}{coefficient \:\: of \:\: x^{2}}
Product of zeroes = (-2) X (-5) = 10 = \frac{10}{1} = \frac{constant \:\: term}{coefficient \:\: of \:\: x^{2}}
Example 3 : Find the zeroes of the polynomial x^{2}
Solution : Recall the identity a^{2} – b^{2}
x^{2} – 3 = (X - \sqrt{3}) (X + \sqrt{3})
So, the value of x^{2}
Therefore, the zeroes of x^{2} – 3
Now,
Sum of zeroes = \sqrt{3} - \sqrt{3} = 0 = \frac{-(coefficient \:\: of \:\: x)}{coefficient \:\: of \:\: x^{2}}
Product of zeroes = (\sqrt{3})(-\sqrt{-3} = -3 = \frac{-3}{1} = \frac{constant \:\: term}{coefficient \:\: of \:\: x^{2}}
Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.
Solution : Let the quadratic polynomial be ax^{2} + bx + c
α + β = – 3 = \frac{-b}{a}
and αβ = 2 = \frac{c}{a}
If a = 1, then b = 3 and c = 2.
So, one quadratic polynomial which fits the given conditions is x^{2}
You can check that any other quadratic polynomial that fits these conditions will be of the form k(x^{2} + 3x + 2)
Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients? Let us consider p(x) = 2x^{3} – 5x^{2}
You can check that p(x) = 0 for x = 4, – 2, \frac{1}{2}
zeroes, these are the zeroes of 2x^{3} – 5x^{2} – 14x + 8
sum of the zeroes = 4 + (-2) + \frac{1}{2} = \frac{5}{2} = \frac{-(-5)}{2} = \frac{-(coefficient \:\: of \:\: x^{2})}{coefficient \:\: of \:\: x^{3}}
product of the zeroes = 4 X (-2) X \frac{1}{2} = -4 = \frac{-8}{2} = \frac{-constant \:\: term}{coefficient \:\: of \:\: x^{3}}
However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have
{4 X (-2)} + {(-2) X \frac{1}{2}} + {\frac{1}{2} X 4}
= -8 – 1 + 2 = -7 = \frac{-14}{2} = \frac{coefficient \:\: of \:\: x}{coefficient \:\: of \:\: x^{3}}
In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax^{3} + bx^{2} + cx + d
α + β + γ = \frac{-b}{a}
αβ + βγ + γα = \frac{c}{a}
α β γ = \frac{-d}{a}
Let us consider an example.
Example 5* : Verify that 3, –1, -\frac{1}{3}
Solution : Comparing the given polynomial with ax^{3} + bx^{2} + cx + d
a = 3, b = – 5, c = –11, d = – 3. Further
p(3) = 3 × 3^{3} – (5 × 3^{2})
p(–1) = 3 × (–1)^{3} – 5 × (–1)^{2} – 11 × (–1) – 3
p(\frac{-1}{3}) = 3 X (-\frac{1}{3})^{3} – 5 X (-\frac{1}{3})^{2} – 11 X (-\frac{1}{3}) – 3
= (-\frac{1}{9} - \frac{5}{9} + \frac{11}{3} – 3 = - \frac{2}{3} + \frac{2}{3}
Therefore, 3, - 1 and -\frac{1}{3}
So, we take α = 3, β = -1 and γ = \frac{1}{3}
α + β + γ = 3 + (-1) + (-\frac{1}{3}) = 2 - \frac{1}{3} = \frac{5}{3} = \frac{-(-5)}{3} = \frac{-b}{a}
αβ + βγ + γα =3 x (-1) + (-1) X (-\frac{1}{3}) + (-\frac{1}{3}) x 3 = -3 + \frac{1}{3} – 1 = \frac{-11}{3} = \frac{c}{a}
αβγ = 3 x (-1) X (-\frac{1}{3}) = 1 = \frac{-(-3)}{3} = \frac{-d}{a}
You know that a cubic polynomial has at most three zeroes. However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial x^{3} – 3x^{2} – x + 3
Next, you could get the factors of x^{2}
x^{3} – 3x^{2}
= (x – 1)(x + 1)(x – 3)
So, all the three zeroes of the cubic polynomial are now known to you as 1, – 1, 3.
Example 6 : Divide 2x^{2}
Solution : Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor. So, here the quotient is 2x – 1 and the remainder is 3. Also,
(2x – 1)(x + 2) + 3 = 2x^{2} + 3x – 2 + 3 = 2x^{2}
Therefore, Dividend = Divisor × Quotient + Remainder
Let us now extend this process to divide a polynomial by a quadratic polynomial.
Example 7 : Divide 3x^{3} + x^{2}
Solution : We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x^{2} + 2x + 1
Step 1 : To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x^{3}
Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e., –5x^{2}
Step 3 : What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x^{2}
So, the quotient is 3x – 5 and the remainder is 9x + 10. Also,
(x^{2}
= 3x^{3} + x^{2}
Here again, we see that
Dividend = Divisor x Quotient + Remainder
If p(x) and g (x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r (x),
where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.
Example 8 : Divide 3x^{2} – x^{3}
Solution : Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees.
So, dividend = –x^{3} + 3x^{2} – 3x + 5
Division process is shown on the right side.
We stop here since degree (3) = 0 < 2 = degree (–x^{2}
So, quotient = x – 2, remainder = 3.
Now,
Divisor × Quotient + Remainder
= (–x^{2}
= –x^{3} + x^{2} – x + 2x^{2} – 2x + 2 + 3
= –x^{3} + 3x^{2} – 3x + 5
= Dividend
Example 9 : Find all the zeroes of 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2
Solution : Since two zeroes are \sqrt{2}
First term of quotient is \frac{2x^{4}}{x^{2}} = 2x^{2}
Second term of quotient is \frac{-3x^{3}}{x^{2}} = -3x
Third term of quotient is \frac{x^{2}}{x^{2}}
So, 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2
Now, by splitting –3x, we factorise 2x^{2}
are given by x = \frac{1}{2}
\sqrt{2}, -\sqrt{2}, \frac{1}{2}
0 Doubts's