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I am doing a course in calculus and I was given this problem :

Given that $f(x)=3x^4−6x^3+4x^2−7x+3$, evaluate $f(−2)$.

The answer is meant to be 129 according to the tutor, but no matter how many times I try and work it out I can't see how they got that answer.

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Please tell us what you tried, and what answer your attempted solution gave. –  Andrew Uzzell Jan 10 '13 at 14:40
    
The answer is correct, and it's just substituting values, what problem are you having? –  MyUserIsThis Jan 10 '13 at 14:42
    
My second term was -48 when it had to be positive. –  8BitSensei Jan 10 '13 at 14:51
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5 Answers 5

up vote 3 down vote accepted

It appears your problem is with order of operations.

Let's look at an easier function: $$g(x) = 6x^4$$

What is $g(-2)$?

$$g(-2) = 6(-2)^4$$

Remember--you evaluate exponents before multiplication: $$g(-2) = 6(16)$$ $$g(-2) = 96$$

If the positive/negative thing is still hard, just turn the exponent into multiplication: $$g(-2) = 6[(-2)(-2)(-2)(-2)]$$ $$g(-2) = 6[(-2)(-2)(4)]$$ $$g(-2) = 6[(-2)(-8)]$$ $$g(-2) = 6(16)$$ $$g(-2) = 96$$

EDIT:

From comments, it appears the error wasn't with raising the negative to an even power, but rather with the second term. The answer has been dealt with in other responses, but I'll include it here for archive purposes:

$$f(x)=3x^4−6x^3+4x^2−7x+3$$ $$f(-2)=3(-2)^4−6(-2)^3+4(-2)^2−7(-2)+3$$ $$f(-2)=3(16)−6(-8)+4(4)−7(-2)+3$$ (Note the minus signs in front of the six and seven. I now multiply out those negatives, which makes them positive.) $$f(-2)=3(16)+6(8)+4(4)+7(2)+3$$ $$f(-2)=48+48+16+14+3$$ $$f(-2)=129$$

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I found that my problem was that my final abstract was 48-48+16+3, when it should have been, 48+48+16+3, still not entirely sure why the second term becomes positive though. –  8BitSensei Jan 10 '13 at 14:54
    
Oh! ok. I added that part to my answer now. –  anorton Jan 10 '13 at 16:09
    
Thank you, the tutorial I was using did not show any working out with the answer, that made it difficult to work out where I was going wrong. –  8BitSensei Jan 10 '13 at 16:12
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Horner's method? $$3 + x \cdot ( -7 + x \cdot (4 + x \cdot (-6 + 3\cdot x)))=$$ $$=3 + (-2 )\cdot ( -7 + (-2) \cdot (4 + (-2) \cdot (-12)))=$$ $$=3 + (-2) \cdot ( -7 + (-2) \cdot (4 + 24))=$$ $$=3 + (-2) \cdot ( -7 + (-2) \cdot 28)=$$ $$=3 + (-2) \cdot ( -7 - 56)=$$ $$=3 + (-2) \cdot (-63)=$$ $$=129$$

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Why do you want to make it so difficult......... –  MSKfdaswplwq Jan 10 '13 at 14:52
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$$f(x)=3x^4−6x^3+4x^2−7x+3=(x+2)(3x^3-12x^2+28x-63)+129$$

$$\implies f(-2)=0(3x^3-12x^2+28x-63)+129=129$$

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Observe the magic of color:$$f(\color{#C00}{x}) = 3\color{#C00}{x}^4 - 6\color{#C00}{x}^3 + 4\color{#C00}{x}^2 - 7\color{#C00}{x} + 3$$Now, we essentially have to replace our colored expressions with numbers.$$f(-2) = 3(-2)^4 - 6(-2)^3 + 4(-2)^3 - 7(-2) + 3$$Now use the order of operations to simplify. For example, $3(-2)^4 = 3\times(-2)^4 = 3\times16 = {48}$.

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Ummm... it's $f(-2)$ The downvote is not mine, but that's probably the reason you got it... –  anorton Jan 10 '13 at 14:44
    
Thank you, turns out that my hyphen key was broken and I was not looking at what I was typing. :-) –  Parth Kohli Jan 10 '13 at 14:45
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$$f(-2)=3\cdot(-2)^4-6\cdot(-2)^3+4\cdot(-2)^2-7\cdot(-2)+3=48+48+16+14+3=129$$

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(−2)3 = 16, so -6⋅16 = -48, how does it become positive? –  8BitSensei Jan 10 '13 at 14:46
    
The first and third term have an even power, so everything will be nonnegative. The second and fourth term have odd powers, but we're subtracting them, so we're subtracting a negative number which makes it positive. –  Clayton Jan 10 '13 at 14:47
    
Also, $(-2)\cdot 3=-6$, not $16$, and $(-2)^3=-8$. –  Clayton Jan 10 '13 at 14:52
    
the 3 is a exponent of (-2), not very clear in my comment. –  8BitSensei Jan 10 '13 at 14:58
    
I thought that might be the case, and that was why I wanted to point out $(-2)^3=-8$. –  Clayton Jan 10 '13 at 14:59
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