Question

Find the polynomial function with roots 1, 7, and -3 of multiplicity 2. (1 point)

Help?? Anyone?!!!?!?!

Answer 1

\[(x-1)(x-7)(x+3)^2\] or whatever you get when you multiply that out

Answer 2

What would be the three of them multiplied? I've only ever multiplied two.

Answer 3

actually 4

Answer 4

\[(x-1)(x-7)(x+3)^2=(x-1)(x-7)(x+3)(x+3)\] etc

Answer 5

How do I multiply 4 together? Do I distribute x from the first one to every single one and so on?

Answer 6

just work them 2 at a time: distribution is a good notion to keep in mind yes

\[(x-1)(x-7)(x+3)(x+3)\]
\[a(x-7)(x+3)(x+3)\]
\[(ax-7a)(x+3)(x+3)\]
\[b(x+3)(x+3)\]
\[(bx+3b)(x+3)\]
\[c(x+3)\]
\[cx+3c\]

Answer 7

working backwards ...
\[cx+3c~:~\text{but c=bx+3b}\]
\[(bx+3b)x+3(bx+3b)\]
\[bx^2+3bx+3bx+9b\]
\[bx^2+6bx+9b~:~\text{but b=ax-7a}\]
\[x^2(ax-7a)+6x(ax-7a)+9(ax-7a)\]
\[ax^3-7ax^2+6ax^2-42ax+9ax-64a\]
\[ax^3-ax^2-33ax-64a\]
...but a = x-1