Find the other zeroes of the polynomial 3x⁴ – 7x³ – 15x² +21x +18 , where two of its zeroes are √3 & - √3.
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Find the other zeroes of the polynomial 3x⁴ – 7x³ – 15x² +21x +18 , where two of its zeroes are √3 & - √3.
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Step-by-step explanation:
To find the other zeroes of the polynomial 3x⁴ - 7x³ - 15x² + 21x + 18, given that two of its zeroes are √3 and -√3, we can use the fact that if α is a zero of the polynomial, then (x - α) is a factor of the polynomial.
Given zeroes:
α₁ = √3
α₂ = -√3
So, the factors corresponding to these zeroes are:
(x - α₁) = (x - √3)
(x - α₂) = (x + √3)
Now, we can find the other two zeroes using the factor theorem.
Step 1: Divide the original polynomial by the factors (x - √3) and (x + √3) using synthetic division or long division.
The division gives us the quotient: 3x² - 4x - 6
Step 2: Now, solve the quadratic equation 3x² - 4x - 6 = 0 to find the remaining two zeroes.
Using the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a
a = 3, b = -4, c = -6
x = [4 ± √((-4)² - 4 * 3 * (-6))] / 2 * 3
x = [4 ± √(16 + 72)] / 6
x = [4 ± √88] / 6
x = [4 ± 2√22] / 6
x = (2/3) ± (√22)/3
So, the other two zeroes of the polynomial are:
α₃ = (2 + √22) / 3
α₄ = (2 - √22) / 3
Therefore, the four zeroes of the polynomial 3x⁴ - 7x³ - 15x² + 21x + 18 are:
√3, -√3, (2 + √22) / 3, and (2 - √22) / 3.
Answer:
THEREFORE, THE FOUR ZERO OF POLYNOMIAL