How do you write a polynomial with zeros: 2, 4 + sqrt5, 4-sqrt5?

1 Answer
mason m
Feb 9, 2016

#x^4-14x^3+67x^2-130x+88#

Explanation:

Note that: if #a# is a zero of a polynomial, then #(x-a)# is a root of the polynomial. Applying this to the zeros at hand, we see that that polynomial is equal to

#(x-2)(x-4)(x-(4+sqrt5))(x-(4-sqrt5))#

We can simplify the first two through distributing. We will get to the second two, which is the more challenging part of this problem, but first distribute the minus sign.

#=(x^2-6x+8)(x-4-sqrt5)(x-4+sqrt5)#

#=(x^2-6x+8)((x-4)-sqrt5)((x-4)+sqrt5)#

Notice that the last two terms are in the form #(a-b)(a+b)#, which equals #a^2-b^2#. Here, #a=x-4# and #b=sqrt5#. This gives us

#=(x^2-6x+8)((x-4)^2-5)#

#=(x^2-6x+8)(x^2-8x+16-5)#

#=(x^2-6x+8)(x^2-8x+11)#

Now, we can distribute these terms. There will be a lot, so hang on tight.

#=overbrace(x^4-6x^3+8x^2)^((x^2-6x+8) * x^2)+overbrace(-8x^3+48x^2-64x)^((x^2-6x+8) * -8x)+overbrace(11x^2-66x+88)^((x^2-6x+8) * 11)#

Combining all the like terms #x^3# with #x^3#, and so on, our result is

#=x^4-14x^3+67x^2-130x+88#

Note that this polynomial can be multiplied by any constant, and the roots will be the same. However, this is the simplest form of the polynomial.

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