Verify that 1, 3 and 4 are the zeroes of the cubic polynomial p(x) = x3- 8x2 - 19x - 12
and then verify the relationship between the zeroes and the coefficients.
Answers
Solution
Given:-
- 1,3,4 aee zeroes of cubic polynomial p(x) = x³ - 8x² - 19x - 12
Prove :-
- These zeroes are cubic polynomial .
Proof:-
We Know,
If a & b are zeroes of equations ax² + bx + c = 0 , then equation will be satisfied for value of x .
That's means, 1,3,4 will satisfied this cubic polynomial .
For, this we keep value of x one by one.
Case 1.
- When, x = 1
==> p(1) = (1)³ - 8 × (1)² - 19 ×(1) - 12
==> p(1) = 1 - 8 - 19 - 12
==> p(1) = 1 - 39
==> p(1) = -38
Case 2.
- When, x = 3
==> p(3) = (3)³ - 8 × (3)² - 19 × 3 - 12
==> p(3) = 27 - 72 - 57 - 12
==> p(3) = 27 - 141
==> p(3) = -117
Case 3.
- When, x = 4
==> p(4) = 4³ - 8 × 4² - 19 × 4 - 12
==> p(4) = 254 - 96 - 76 - 12
==> p(4) = 254 - 184
==> p(x) = 70
Here, for value of x ,
This cubic polynomial not satisfied by any value of x .
In this case ,
We can say, these zeroes are not this cubic polynomial .
___________________
Answer:
Given:-
1,3,4 aee zeroes of cubic polynomial p(x) = x³ - 8x² - 19x - 12
Prove :-
These zeroes are cubic polynomial .
Proof:-
We Know,
If a & b are zeroes of equations ax² + bx + c = 0 , then equation will be satisfied for value of x .
That's means, 1,3,4 will satisfied this cubic polynomial .
For, this we keep value of x one by one.
Case 1.
When, x = 1
==> p(1) = (1)³ - 8 × (1)² - 19 ×(1) - 12
==> p(1) = 1 - 8 - 19 - 12
==> p(1) = 1 - 39
==> p(1) = -38
Case 2.
When, x = 3
==> p(3) = (3)³ - 8 × (3)² - 19 × 3 - 12
==> p(3) = 27 - 72 - 57 - 12
==> p(3) = 27 - 141
==> p(3) = -117
Case 3.
When, x = 4
==> p(4) = 4³ - 8 × 4² - 19 × 4 - 12
==> p(4) = 254 - 96 - 76 - 12
==> p(4) = 254 - 184
==> p(x) = 70
Here, for value of x ,
This cubic polynomial not satisfied by any value of x .
In this case ,
We can say, these zeroes are not this cubic polynomialGiven:-
1,3,4 aee zeroes of cubic polynomial p(x) = x³ - 8x² - 19x - 12
Prove :-
These zeroes are cubic polynomial .
Proof:-
We Know,
If a & b are zeroes of equations ax² + bx + c = 0 , then equation will be satisfied for value of x .
That's means, 1,3,4 will satisfied this cubic polynomial .
For, this we keep value of x one by one.
Case 1.
When, x = 1
==> p(1) = (1)³ - 8 × (1)² - 19 ×(1) - 12
==> p(1) = 1 - 8 - 19 - 12
==> p(1) = 1 - 39
==> p(1) = -38
Case 2.
When, x = 3
==> p(3) = (3)³ - 8 × (3)² - 19 × 3 - 12
==> p(3) = 27 - 72 - 57 - 12
==> p(3) = 27 - 141
==> p(3) = -117
Case 3.
When, x = 4
==> p(4) = 4³ - 8 × 4² - 19 × 4 - 12
==> p(4) = 254 - 96 - 76 - 12
==> p(4) = 254 - 184
==> p(x) = 70
Here, for value of x ,
This cubic polynomial not satisfied by any value of x .
In this case ,
We can say, these zeroes are not this cubic polynomial