Answer:
To find a cubic polynomial with the given information, we can start by assuming the polynomial in the form:
P(x) = (x - a)(x - b)(x - c)
Where a, b, and c are the zeroes of the polynomial.
Given the product of the zeroes is -14, we have:
abc = -14
Given the sum of the product of the zeroes taken two at a time is -7, we have:
ab + ac + bc = -7
Given the product of the zeroes is 2, we have:
abc = 2
Now, we can solve these equations to find the values of a, b, and c.
From the equation abc = -14, we can rewrite it as c = -14/(ab).
Substituting this value in the equation ab + ac + bc = -7, we get:
ab + a(-14/(ab)) + b(-14/(ab)) = -7
ab^2 - 14a - 14b = -7ab
Multiplying the entire equation by ab, we have:
a^2b^2 - 14ab - 14ab^2 = -7ab^2
Rearranging terms, we get:
a^2b^2 + 7ab^2 - 14ab - 14ab^2 = 0
Factoring out common terms, we have:
ab(a^2 + 7b - 14 - 14b) = 0
Setting each factor equal to zero, we have two possibilities:
ab = 0 (which implies either a = 0 or b = 0)
a^2 + 7b - 14 - 14b = 0
If ab = 0, we can consider two cases:
Case 1: a = 0
From abc = 2, we have 0bc = 2, which is not possible since the product of the zeroes is given as 2.
Case 2: b = 0
From abc = 2, we have a0c = 2, which is also not possible.
Therefore, we can conclude that ab ≠ 0.
Now, let's solve the equation a^2 + 7b - 14 - 14b = 0 for the values of a and b.
a^2 + (7 - 14)b - 14 = 0
a^2 - 7b - 14 = 0
Using the quadratic formula:
a = [ -(-7) ± √((-7)^2 - 4(1)(-14)) ] / (2(1))
a = [ 7 ± √(49 + 56) ] / 2
a = [ 7 ± √(105) ] / 2
Hence, we have two possible values for a:
a1 = (7 + √105) / 2
a2 = (7 - √105) / 2
Now, substituting these values in the equation abc = 2, we can find the corresponding values of c.
For a1:
a1b*c = 2
[(7 + √105) / 2] * b * c = 2
b * c = (2 * 2) / (7 + √105)
b * c = 4 / (7 + √105)
For a2:
a2b*c = 2
[(7 - √105) / 2] * b * c = 2
b * c = (2 * 2) / (7 - √105)
b * c = 4 / (7 - √105)
Now, we have multiple valid combinations for the values of b and c that satisfy the equation.
One possible set of values for b and c are:
b = 2
c = 4 / (7 + √105)
Thus, we have found one cubic polynomial with the given zeroes:
P(x) = (x - a1)(x - b)(x - c)
P(x) = (x - (7 + √105) / 2)(x - 2)(x - 4 / (7 + √105))
Hope it helps
Step-by-step explanation:
I HOPE IT'S HELPFUL.......
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Answers & Comments
Answer:
To find a cubic polynomial with the given information, we can start by assuming the polynomial in the form:
P(x) = (x - a)(x - b)(x - c)
Where a, b, and c are the zeroes of the polynomial.
Given the product of the zeroes is -14, we have:
abc = -14
Given the sum of the product of the zeroes taken two at a time is -7, we have:
ab + ac + bc = -7
Given the product of the zeroes is 2, we have:
abc = 2
Now, we can solve these equations to find the values of a, b, and c.
From the equation abc = -14, we can rewrite it as c = -14/(ab).
Substituting this value in the equation ab + ac + bc = -7, we get:
ab + a(-14/(ab)) + b(-14/(ab)) = -7
ab^2 - 14a - 14b = -7ab
Multiplying the entire equation by ab, we have:
a^2b^2 - 14ab - 14ab^2 = -7ab^2
Rearranging terms, we get:
a^2b^2 + 7ab^2 - 14ab - 14ab^2 = 0
Factoring out common terms, we have:
ab(a^2 + 7b - 14 - 14b) = 0
Setting each factor equal to zero, we have two possibilities:
ab = 0 (which implies either a = 0 or b = 0)
a^2 + 7b - 14 - 14b = 0
If ab = 0, we can consider two cases:
Case 1: a = 0
From abc = 2, we have 0bc = 2, which is not possible since the product of the zeroes is given as 2.
Case 2: b = 0
From abc = 2, we have a0c = 2, which is also not possible.
Therefore, we can conclude that ab ≠ 0.
Now, let's solve the equation a^2 + 7b - 14 - 14b = 0 for the values of a and b.
a^2 + (7 - 14)b - 14 = 0
a^2 - 7b - 14 = 0
Using the quadratic formula:
a = [ -(-7) ± √((-7)^2 - 4(1)(-14)) ] / (2(1))
a = [ 7 ± √(49 + 56) ] / 2
a = [ 7 ± √(105) ] / 2
Hence, we have two possible values for a:
a1 = (7 + √105) / 2
a2 = (7 - √105) / 2
Now, substituting these values in the equation abc = 2, we can find the corresponding values of c.
For a1:
a1b*c = 2
[(7 + √105) / 2] * b * c = 2
b * c = (2 * 2) / (7 + √105)
b * c = 4 / (7 + √105)
For a2:
a2b*c = 2
[(7 - √105) / 2] * b * c = 2
b * c = (2 * 2) / (7 - √105)
b * c = 4 / (7 - √105)
Now, we have multiple valid combinations for the values of b and c that satisfy the equation.
One possible set of values for b and c are:
b = 2
c = 4 / (7 + √105)
Thus, we have found one cubic polynomial with the given zeroes:
P(x) = (x - a1)(x - b)(x - c)
P(x) = (x - (7 + √105) / 2)(x - 2)(x - 4 / (7 + √105))
Hope it helps
Step-by-step explanation:
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Answer:
I HOPE IT'S HELPFUL.......