Find a polynomial function f(x) with real coefficients that has the given zeroes. 0,6,1+√2i

Answer 1

Zeros: 0, 6, and 1 + √2 i,

Now, because we know that radical solutions and imaginary solutions come in pairs, we also know that 1 - √2 i is also a solution. So our zeros really are

Zeros: 0, 6, 1 ± √2 i

Let's set up our function:

f(x) = (x - 0)(x - 6[(x - (1 + √2 i)][x - (1- √2 i)]

= x(x - 6)(x - 1 - √2 i)(x - 1 + √2 i) <-------Distribute the negative inside the last two factors.

Now this is where it gets interesting. The last two terms can be a painful one, so we group them as follows:

[(x - 1) - √2 i][(x - 1) + √2 i] <------The reason for this grouping is so that we create the pattern (a + b)(a - b) Do you see it? Here a = (x - 1) and b = √2 i. Thus, when expanded we just need to do a² - b²

(x - 1)² - 2i²

= (x² - 2x + 1) + 2

= x² - 2x + 3

Thus,

x(x - 6)(x - 1 - √2 i)(x - 1 + √2 i)

= x(x - 6)(x² - 2x + 3)

= x(x³ - 2x² + 3x - 6x² + 12x - 18)

= x(x³ - 8x² + 15x - 18)

= x⁴ - 8x³ + 15x² - 18x

As you can see it's a whole lot of algebra, but overall, not too difficult. You will have to be very careful on these types of problems, one mistake and it's all over. My suggestion: Do the multiplication slowly.

Hope this helps

Mr. K