The zeros of a polynomial p(x) are all the x-values that make the polynomial equal to zero.
What is the meaning of zeroes ?
By finding zeroes of the polynomial, we can understand that where the curve is intersecting the x-axis.
How many zeroes will be there for a polynomial ?
Based on the highest exponent, we can decide number of zeroes that we will receive.
Example 1 :
Find zero of the polynomial p(x) = x + 3.
Solution :
The given polynomial p(x) is a linear polynomial, the number of zeroes for this polynomial will be 1.
p(x) = 0
x + 3 = 0
x = -3
-3 is the zero of the polynomial.
Example 2 :
Find the zeroes of quadratic polynomial x²-5x-6.
Solution :
Since the given is quadratic polynomial, it will have two zeroes.
Let p(x) = x²-5x-6
0 = x²-5x-6
(x - 6) (x + 1) = 0
x = 6 and x = -1
p(x) = x²-5x-6 if x = 6 p(6) = 6²-5(6)-6 = 36 - 30 - 6 = 36 - 36 p(6) = 0 |
p(x) = x²-5x-6 if x = -1 p(-1) = (-1)²-5(-1)-6 = 1 + 5 - 6 = 6 - 6 p(-1) = 0 |
So, 6 and -1 are the zeroes of the polynomial.
Example 3 :
If 1 is a zero of the polynomial p(x) = ax^{2} - 3(a-1)x - 1, then find the value of 'a' ?
Solution :
p(x) = ax^{2} - 3(a-1)x - 1
1 is a zero of the polynomial, then p(1) = 0
p(x) = ax^{2} - 3(a-1)x - 1
p(1) = a(1)^{2} - 3(a-1)(1) - 1 = 0
a - 3a + 3 - 1 = 0
-2a + 2 = 0
-2a = -2
a = -2/(-2)
a = 1
Example 4 :
If the graph of a polynomial intersects the x – axis at only one point, can it be a quadratic polynomial?
Solution :
When we solve quadratic polynomial, we will be having two values.
There are three possible solutions.
1) It may have same zeroes, so the curve will intersect the x-axis at one point.
2) It may have two different zeroes, so the curve will intersect at two different points of x.
3) It may not have real values as zeroes, so the curve will not intersect the x-axis.
So, the result for the above question is true.
Example 5 :
What number should be added to the polynomial x^{2}-5x+4, so that 3 is the zero of the polynomial?
Solution :
Since 3 is the zero,
p(x) = x^{2}-5x+4
p(3) = 3^{2 }- 5(3) + 4
p(3) = 9 - 15 + 4
p(3) = 13 - 15
p(3) = -2
So, the given polynomial should be added with 2.
Example 6 :
If x – 3 and x – 1/3 are both factors of ax^{2} + 5x + b , show that a = b
Solution :
x - 3 = 0, then x = 3
x - 1/3 = 0, then x = 1/3
Let p(x) = ax^{2} + 5x + b
For p(3) and p(1/3), we will get 0.
p(3) = a(3)^{2} + 5(3) + b p(3) = 9a + 15 + b 0 = 9a + 15 + b 9a + b = -15 --(1) |
p(1/3) = a(1/3)^{2} + 5(1/3) + b p(1/3) = a/9 + 5/3 + b 0 = a/9 + 5/3 + b 0 = a + 15 + 9b a + 9b = -15 ----(2) |
(1) - (2)
(9a + b) - (a + 9b) = -15 - (-15)
9a - a + b - 9b = -15 + 15
8a - 8b = 0
a - b = 0
a = b
Hence it is proved.
Example 7 :
If y = -1 is a zero of the polynomial q(y) = 4y^{3} + ky^{2} - y -1, then find the value of k
Solution :
q(y) = 4y^{3} + ky^{2} - y -1
Since -1 is the zero of the polynomial q(-1) = 0
q(-1) = 4(-1)^{3} + k(-1)^{2} - (-1) -1
0 = 4(-1) + k(1) + 1 - 1
0 = -4 + k
k = 4
Example 8 :
For what value of m is x^{3} – 2mx^{2} + 16 divisible by x + 2
Solution :
Let p(x) = x^{3} – 2mx^{2} + 16
Since it is divisible by x + 2.
x + 2 = 0
x = -2
p(-2) = (-2)^{3} – 2m(-2)^{2} + 16
0 = -8 - 2m(4) + 16
0 = 8 - 8m
8m = 8
m = 1
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May 21, 24 08:51 AM
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