Given that√2 is the root of cubic polynomial 6x3 + 2x2-10x -4√2 find the other two roots
f(x)=6x3+√2x2-10x-4√2
Since x=√2 is a zero of f(x)
=>(x-√2) is a factor of f(x)
x-√2)6x3+√2x2-10x-4√2(6x2+7√2x+4
6x3-6√2x2
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7√2x2-10x-4√2
7√2x2-14x
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4x-4√2
4x-4√2
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O
Now, 6x2+7√2x+4=0
=> 6x2+4√2x+3√2x+4=0
=> 2x(3x+2√2)+√2(3x+2√2)=0
=> (3x+2√2)(2x+√2)=0
=> 3x+2√2=0 or 2x+√2=0
=> x=-2√2/3 or x=-√2/2
a little synthetic division shows that (x-√2) does not divide the cubic.
In fact f(√2) = 4-2√2
Check for typos and try again.
Once you have done the division, you will be left with a quadratic, which you can then solve in the usual way.
To find the other two roots of the cubic polynomial, we can use the fact that if α is a root of a polynomial, then its conjugates are also roots of the polynomial.
We are given that √2 is one of the roots of the cubic polynomial 6x^3 + 2x^2 - 10x - 4√2.
Step 1: Find the conjugate of √2.
The conjugate of a square root expression is just the negative of the expression. Therefore, the conjugate of √2 is -√2.
Step 2: Knowing the conjugate, we can find the other two roots.
Since √2 is one of the roots, we have:
(x - √2) as a factor.
To find the other two roots, we divide the cubic polynomial by (x - √2):
(6x^3 + 2x^2 - 10x - 4√2) ÷ (x - √2)
Performing polynomial division, we get:
____________________
(x - √2) │ 6x^3 + 2x^2 - 10x - 4√2
6x^2 + 8√2x + 16
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x - √2 │ 6x^3 + 2x^2 - 10x - 4√2
-(6x^3 - 6√2x^2)
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8√2x^2 - 10x
-(8√2x^2 - 8√2x)
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-2√2x - 4√2
-( -2√2x + 2√2)
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-6√2
The quotient after performing division is 6x^2 + 8√2x + 16. Therefore, our cubic polynomial can be factored as:
6x^3 + 2x^2 - 10x - 4√2 = (x - √2)(6x^2 + 8√2x + 16)
Now we can solve the quadratic factor 6x^2 + 8√2x + 16 to find the other two roots.
Using the quadratic formula,
x = [-b ± √(b^2 - 4ac)] / 2a
In our case, a = 6, b = 8√2, and c = 16.
Applying these values to the formula:
x = [-8√2 ± √((8√2)^2 - 4*6*16)] / (2*6)
x = [-8√2 ± √(128 - 384)] / 12
x = [-8√2 ± √(-256)] / 12
x = [-8√2 ± 16i] / 12
x = (-2√2 ± 4i) / 3
Therefore, the other two roots are (-2√2 + 4i) / 3 and (-2√2 - 4i) / 3.
To find the other two roots of the cubic polynomial, we can use the concept of synthetic division to divide the polynomial by (x - √2).
Step 1: Write the cubic polynomial in descending order of powers:
6x^3 + 2x^2 - 10x - 4√2
Step 2: Set up synthetic division:
√2 | 6 2 -10 -4√2
|___________________
Step 3: Perform synthetic division by following these steps:
a) Bring down the coefficient of the highest power term (6 in this case).
b) Multiply the divisor (√2) by the quotient value, and write the result below the next coefficient (-4√2 * √2 = -4 * 2 = -8)
c) Add the newly obtained value (-8) to the corresponding coefficient.
d) Repeat steps b and c until all coefficients are processed.
Here's the synthetic division:
√2 | 6 2 -10 -4√2
|___________________
6√2 - 8
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6 6√2 -18 -4√2
Step 4: Simplify the result after synthetic division:
The simplified equation after synthetic division is:
6x^2 + 6√2x - 18 - 4√2 = 0
Step 5: Factorize the simplified equation:
Let's factorize the equation:
6x^2 + 6√2x - 18 - 4√2 = 0
6(x^2 + √2x - 3 - 2√2) = 0
Step 6: Solve for the remaining two roots:
Now, equate each factor (x^2 + √2x - 3 - 2√2) to 0 and solve for x:
x^2 + √2x - 3 - 2√2 = 0
To find the roots of this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation (x^2 + √2x - 3 - 2√2 = 0):
a = 1, b = √2, and c = -3 - 2√2
x = (-(√2) ± √((√2)^2 - 4(1)(-3 - 2√2))) / (2(1))
Simplifying further:
x = (-√2 ± √(2 - 4(-3 - 2√2))) / 2
x = (-√2 ± √(2 + 12 + 8√2)) / 2
x = (-√2 ± √(14 + 8√2)) / 2
x = (-√2 ± √(14) ± √(8√2)) / 2
x = (-√2 ± 2√2 ± √(8√2)) / 2
x = (-√2 ± √2(1 + 2) ± √(8√2)) / 2
x = (-√2 ± √2(3) ± 2√2) / 2
x = (-√2 ± √2(3 + 2)) / 2
x = (-√2 ± √2(5)) / 2
x = (-√2 ± √(4)) / 2
x = (-√2 ± 2) / 2
Therefore, the other two roots of the cubic polynomial 6x^3 + 2x^2 - 10x - 4√2 are:
(-√2 + 2) / 2
(-√2 - 2) / 2
So the two other roots are (-√2 + 2)/2 and (-√2 - 2)/2.