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Ad libitum [116K]
3 years ago
7

If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.

Mathematics
2 answers:
Sophie [7]3 years ago
8 0
There's no if about it, 

f(x)=x^3+3x^2-x-3


has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

max2010maxim [7]3 years ago
7 0

I think you got it backwards! Its <em>X=-3, X=-1, or X=1.</em> So the answer is B. (2nd option) Well, you got it! ;)

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GenaCL600 [577]
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3 years ago
HELP
Sonbull [250]

Answer:

A. 121 ⇒ III. 11

B. 64 ⇒ II. 4 and IV. 8

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D. 125 ⇒ V. 5

E. 16 ⇒ II. 4

Step-by-step explanation:

Let us find the correct answer

∵ 121 = 11 × 11

∴ The square root 121 is 11

∴ A. 121 ⇒ III. 11

∵ 64 = 8 × 8

∴ The square root of 64 is 8

∵ 64 = 4 × 4 × 4

∴ The cube root of 64 is 4

∴ B. 64 ⇒ II. 4 and IV. 8

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5 0
3 years ago
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