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Suppose that $x^*$ means $5x^2 - x$. Then what does $(y+3)^*$ mean?

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Are there parenthesis missing? Is that really $5\times 2$? If so, why not write $10$ instead? –  Matt Sep 5 '12 at 13:35
    
@hardmath: Avoid editing the question until the OP has clarified the intended meaning. –  Brandon Carter Sep 5 '12 at 13:38
    
5x^2 is correct. –  Ctrl Sep 5 '12 at 13:39

2 Answers 2

up vote 2 down vote accepted

If $x^* = 5 x^2 - x$ then

$$(y+3)^* = 5(y+3)^2 - (y+3 )= 5(y^2 + 6 y + 9) - y - 3 = 5 y^2 + 29 y + 42 $$

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There's a sign error in the last expression (the x coefficient is net positive), but one character edits are not allowed! I leave it to your discretion. –  hardmath Sep 5 '12 at 16:00
    
Thanks. Edited. –  vanna Sep 5 '12 at 16:01
    
So, +42 is correct? –  Ctrl Sep 5 '12 at 19:03
    
Yes. You can quickly check if this is correct by substituting $y$ to a particular value and look at what you get on the left-hand side and on the right-hand side. Try $y=-3$ you get $0$ on left and $5\times 9-29\times 3+42 = 0$ on right ;) –  vanna Sep 6 '12 at 8:35

By 5x2 do you mean $5x^2$? In such a case, simply replace every instance of $x$ with $(y+3)$, and then expand.

For example, $f(x) = 2x+2 \Longrightarrow f(y+3) = 2(y+3)+2 = 2y+6+2 = 2y+8$.

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