Online Course Discussion Forum

MC 2A Help 2

Hey David, here are some hints for problems 26 - 29:

26: Use substitution. Note the left side of the equation is $x + xy + y = (x+y) + xy$ and the right side of the equation is $x^2 + xy +y^2 = (x+y)^2 -xy$.

27: Part (a) says we can think of $x^3 + y^3 + z^3 -3xyz$ as a polynomial in $z$, that is, think of $x$ and $y$ as constants. So, for part (b) you want to plug in $z = -(x+y)$ and verify that the expression you get is equal to $0$. Then, in (c), since $z = -(x+y)$ is a root of the polynomial, it must be true that $z - (-(x+y)) = x + y + z$ is a factor of $x^3 + y^3 + z^3 -3xyz$, so you can use long division to find the other factor.

28: Try using the general version of Vieta's Theorem. For example, you can start by noting that $t = -(a+b)(b+c)(c+a)$ since $a+b$, $b+c$, and $c+a$ are the roots of the second polynomial.

29: Try rewriting the expression using $u = x+y$ and $v = xy$.

Hope this helps!

28) contains imaginary numbers. How is that taken care of?

Vieta's formulas are true regardless if the roots are real or imaginary. You do not want to find the roots, you just want to find the value of $t$, so there's no need to worry if the roots are real or imaginary.