PreCalc & Trig help? Find a polynomial with a zero at 3 and a zero of multiplicity 2 at -1. Need to show work to get full credit, not really sure how to do this.
If a polynomial has a zero at 3, then one of the factors of the polynomial must equal zero when x = 3. You can deduce what the factor is from that. For the zero at -1, it's a similar process, only there are two factors which equal zero when x = -1, meaning that one of the factors of the polynomial will be to the power of 2. After you've got all your factors, multiply the factors together and simplify to get your polynomial.
a "zero" is a number that makes the value of the polynomial equal to zero. they are also called roots.
multiplying is simple to get a value of zero. so long as one of the numbers you are multiplying is a zero, your set
okay, @Fruitbasket i did something similar, i think, I got it down to be, \[x ^{3}-x ^{2}-5x-3\] Is that correct?
oops, or @amistre64 , sorry didnt see your post until just now
'sok, my posts are really just my recollections :)
Yeah, that's right :)
Come to think of it, you may have to put an = 0 at the end of it.
awesome! thank you! can one of you help me with this one? I'm really lost on what to do.
3^3-3^2-5(3)-3 27 - 9 - 15 - 3 18 - 18 = 0 (-1)^3-(-1)^2-5(-1)-3 -1-1+5 -3 -2 + 2 = 0 .... does seem good
do you know the formula for the volume of a box?
Try to find expressions for the height, width, and length of the box. The function V for the volume is just height x width x length.
so the length and width are the same, because it's square. i think they are both (21-x), which would mean that the volume is (21-x)(21-x)(x) or \[441x-x ^{3}\] is that close?
You've got the right idea, but look at the diagram again. The width and length are the same as you said, but they're not 21-x.
would they be (21+x) ?
x^2 it's either 21-x^2 or 21+x^2
|dw:1365001095235:dw| The sides become smaller, but there are two sections of length x that are taken away.
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