MHB What Mistake Was Made in This Polynomial Long Division?

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The discussion focuses on the polynomial long division of the expression (x^3 + 4x^2 + 3) by (x + 4). The initial attempt incorrectly handled the subtraction step, leading to an erroneous result. The correct process shows that after dividing, the quotient is x^2 with a remainder of 3, resulting in the final answer of x^2 + 3/(x + 4). The mistake was identified in the subtraction of terms during the division process. Proper execution of polynomial long division yields the correct quotient and remainder.
shamieh
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rewrite using polynomial long division

$$\frac{x^3 + 4x^2 + 3}{x+4}$$

so I did $$x+4 \sqrt{x^3 + 4x^2 + 0x + 3}$$

and got $$x^2 + 1 - \frac{1}{x+4}$$

What am i doing wrong? How is that incorrect?
 
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I think you made an error in your first subtraction:

$x^3 + 4x^2 + 3 - (x + 4)x^2 = x^3 + 4x^2 + 3 - x^3 - 4x^2 = 3$.
 
Hello, shamieh!

Rewrite using polynomial long division: $$ \;\;\frac{x^3 + 4x^2 + 3}{x+4}$$

[math]\begin{array}{cccccccccc}
&&&& x^2 & + & 0x & + & 0 \\
&& --&--&--&--&--&--&-- \\
x+4 & ) & x^3 & + & 4x^2 &+&0x& + & 3 \\
&& x^3 & + & 4x^2 \\
&& --&--&-- \\
&&&& 0x^2 &+& 0x \\
&&&& 0x^2 &+& 0x \\
&&&& --&--&--&--\\
&&&&&& 0x &+& 3 \\
&&&&&& 0x &+& 0 \\
&&&&&& --&--&-- \\
&&&&&&&& 3 \end{array}[/math]
Answer: [math]\;x^2 + \frac{3}{x+4}[/math]
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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