since the coefficients are rational, any irrational roots come in conjugate pairs, so another root is 1+√2
(x-(1-√2))(x-(1+√2)) = x^2-2x-1
So, dividing by that we have
(−4x^3 + 6x^2 + 8x + 2)/(x^2-2x-1) = -4x-2 with no remainder
So,
f(x) = (-4x-2)(x-(1+√2))(x-(1-√2)) = (x-(1-√2))(-4x^2+(2+4√2)x+2+2√2) + 0
Write the function in the form
f(x) = (x − k)q(x) + r
for the given value of k.
f(x) = −4x^3 + 6x^2 + 8x + 2, k = 1 − _/2
(square root of 2 if you didn't get what I meant I think _/ looks a bit like the square root symbol so I used it to represent it)
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