Question

Write a polynomial function in standard form with the zeros: -1,1,0.
Please explain steps, don't just give answer.
*Medal to best answer*

Answer 1

If we have a polynomial \(P(x)\) with zeros \(r_1,r_2,...r_n\) we can write it as a product of binomials \[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is an arbitrary constant if we need to have the function go through a specific point, or 1 if we don't care.

To put it in standard form, you multiply it out, collect like terms, and arrange in descending order of the exponents: \(ax^3+bx^2+cx + d\)

Answer 2

No idea what you're talking about.

Answer 3

So, if we had to make a polynomial function with the zeros 2 and -2, we would write
\[P(x) = 1(x-2)(x+2) = x*x + 2x - 2x - 2*2 \]\[P(x)= x^2 -4 \]
Testing it out:
\[P(2) = (2)^2-4 = 4-4=0\checkmark\]\[P(-2) = (-2)^2-4 = 4-4=0\checkmark\]

Answer 4

It's just filling in the blanks, and doing a polynomial multiplication.

Answer 5

Do you understand what it means for a polynomial to have a zero?

Answer 6

No, I don't understand anything because I'm sick and I missed school so I got notes but not a lecture and I'm no good at teaching myself (as you can see). There was one example she did and it showed FOILing. (multiplying First, Outside,Inside, Last) But this is a 3 equation thing and I'm used to only doing two.

Answer 7

Well, what would you like me to explain first? How to multiply 3 binomials together, or what a zero means, or something else entirely?

Answer 8

I guarantee that I can successfully explain this to you, or you'll get a full refund of every last cent you've paid me :-)

Answer 9

How to multiply 3 binomials.

Answer 10

Okay. First thing to do is to throw away the FOIL crutches and stand tall :-)

\[(x+a)(x+b)\]

When you use FOIL, you multiply First
\[(x+a)(x+b) = x^2\]
Outside\[(x+a)(x+b)=x^2+x*b\]
Inside\[(x+a)(x+b)=x^2+x*b+a*x\]
Last\[(x+a)(x+b)=x^2+x*b+a*x+a*b\]and then collect like terms\[(x+a)(x+b)=x^2+x*b+a*x+a*b = x^2 + (a+b)x + a*b\]
Right?

Answer 11

What you are really doing is applying the distributive property repeatedly:
\[(x+a)(x+b) = x(x+b) + a(x+b)\]\[(x+a)(x+b) = x(x+b) + a(x+b) = x*x + b*x + a*x + a*b\]\[(x+a)(x+b) = x^2 + (a+b)x + a*b\]

Answer 12

And in general, you could describe the process as
"take each part of the first thing, multiply it by every individual part of the second thing, and add up all the results"

I say that you should think of it this way because it won't be long before you are multiplying things other than binomials, and then the FOIL method doesn't work, because you have extra pieces. For example:
\[(x+1)(x+2) = x^2 + 2x+1x + 1*2 = x^2 + 3x + 2\]What if we need to now multiply that by \((x+3)\)?
\[(x+3)(x^2+3x+2)\]How do we use FOIL on that?

We don't. Instead, we use the distributive property as many times as is necessary:
\[(x+3)(x^2+3x+2) = x(x^2+3x+2) + 3(x^2+3x+2)\]\[=x*x^2 + x*3x + x*2 + 3(x^2+3x+2)= x^3+3x^2+2x + 3(x^2+3x+2)\]\[=x^3+3x^2+2x+3*x^2+3*3x+3*2\]\[=x^3+3x^2+2x+3x^2+9x+6\]Now we collect all the like terms:\[=x^3+6x^2+11x+6\]

As you can see if you look carefully, we took each piece of the first binomial and multiplied it by each part of the second thing (the trinomial).

Answer 13

So, we just multiplied 3 binomials:\[(x+1)(x+2)(x+3)\]We did so by multiplying the first two, then multiplying the product of the first two by the third. Had we done it all in one big lump, it would have looked like this:
\[(x+1)(x+2)(x+3) = (x^2 + 2x + 1x + 1*2)(x+3) = (x^2+3x+2)(x+3) \]\[ = (x+3)(x^2+3x+2)\]\[=x*x^2 + x*3x+x*2 + 3*x^2 + 3*3x + 3*2 = x^3 + 3x^2+2x + 3x^2 + 9x + 6\]\[=x^3 + 6x^2 + 11x + 6\]

Answer 14

[crickets]