Given that^2 is a zero of the cubic polynomial 6x3+2x2-10x-4^2, find its other two zeroes
what is ^2 ?
Do you mean √2 ?
If so, then you have
6x^3+2x^2-10x-4√2
That does not have x-√2 as a factor.
So, enlighten us as to your notation.
To find the other two zeroes of the cubic polynomial, we need to use the fact that one of the zeroes is given as (4^2)^2, which simplifies to 16^2.
First, let's set up the polynomial equation using the given zero. We know that the polynomial has a factor of (x - (4^2))^2 = (x - 16)^2.
Now we can divide the original cubic polynomial by this factor to obtain a quadratic polynomial, which will help us in finding the other two zeroes.
Performing the long division, we have:
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(x - 16)^2 | 6x^3 + 2x^2 - 10x - 4
- (6x^3 - 96x^2)
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98x^2 - 10x - 4
We now have a quadratic polynomial, 98x^2 - 10x - 4. To find the remaining zeroes, we can now solve this quadratic equation.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 98, b = -10, and c = -4.
Calculating:
x = [ -(-10) ± √((-10)^2 - 4 * 98 * (-4)) ] / (2 * 98)
x = [ 10 ± √(100 + 1568) ] / 196
x = [ 10 ± √1668 ] / 196
To simplify further, we can simplify the square root of 1668. We find that √1668 = 2√417.
Substituting back into the formula:
x = [ 10 ± 2√417 ] / 196
Therefore, the two remaining zeroes of the cubic polynomial are (10 + 2√417) / 196 and (10 - 2√417) / 196.