To find zeroes of a polynomial, we have to equate the polynomial to zero and solve for the variable.
To check whether 'k' is a zero of the polynomial f(x), we have to substitute the value 'k' for 'x' in f(x).
If f(k) = 0, then 'k' is a zero of the polynomial f(x).
Find the zeros of the following polynomials
Question 1 :
p(x) = 4x-1
Solution :
Let p(x) = 4x-1
4x-1 = 0
4x = 1
x = 1/4
So, 1/4 is the zero of the given polynomial 4x-1.
Question 2 :
p(x) = 3x+5
Solution :
Let p(x) = 3x+5
3x+5 = 0
3x = -5
x = -5/3
So, -5/3 is the zero of the given polynomial 3x+5.
Question 3 :
p(x) = 2x
Solution :
Let p(x) = 2x
2x = 0
x = 0/2
x = 0
So, 0 is the zero of the given polynomial 2x.
Question 4 :
p(x) = x+9
Solution :
Let p(x) = x+9
x+9 = 0
x = -9
So, -9 is the zero of the given polynomial x+9.
Verify whether the following are roots of the polynomial equations indicated against them.
Question 1 :
x2 - 5x +6 = 0 x = 2, 3
Solution :
Given that p(x) = x² - 5x +6
p(2) = 22 - 5(2) + 6
= 4 -10 + 6
p(2) = 0
Given that p(x) = x2 - 5x +6
p(3) = 32-5(3)+6
= 9 -15 + 6
p(3) = 0
So, 2 and 3 are zeroes or roots of the polynomial p(x).
Question 2 :
x2 +4x+3 = 0 : x = -1, 2
Solution :
Given that p(x) = x2 +4x+3
p(-1) = (-1)2+ 4(-1) +3
= 1-4+3
p(-1) = 0
p(2) = 22+4(2)+3
= 4+8+3
p(2) = 11 ≠ 0
So, -1 is a zero of the polynomial and 2 is not a zero.
Question 3 :
x3 -2x2-5x+6 = 0 : x = 1, -2, 3
Solution :
Given that p(x) = x3 -2x2-5x+6
p(-1) = x3 -2x2-5x+6
p(1) = (1)3 -2(1)2-5(1) +6
= 1-2-5+6
p(-1) = 0
p(-2) = (-2)3 -2(-2)2-5(-2) +6
= -8-8+10+6
= -16+16
p(-2) = 0
p(3) = 33 -2(3)2-5(3) +6
= 27-18-15+6
= 33-33
p(3) = 0
So, -1, -2 and 3 are zeroes or roots of the given polynomial.
Question 4 :
x3-2x2-x+2 = 0: x = -1, 2
Solution :
Given that p(x) = x3-2x2-x+2
p(-1) = x3-2x2-x+2
p(-1) = (-1)3 -2(-1)2-(-1)+2
= -1-2+1+2
= -3+3
p(-1) = 0
p(2) = 23 -2(2)2-2+2
= 8-2(4)-2+2
= 8-8-2+2
p(2) = 0
So, -1 and 2 are zeroes of the polynomial.
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