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The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.

Please give an explanation of how the answer

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    $\frac{a x^3+b x^2+25}{x^2-4x+5}=4a+b+a x+\frac{(11a+4b)x-5(4a+b)+25}{x^2-4x+5}$2011-01-04

4 Answers 4

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Since $x^2-4x+5=0$ has the roots $2 \pm i$ they must also be roots of $ax^3+bx^2+25\,$ therefore:

$ 0 = a(2+i)^3+b(2+i)^2+25=8a -6a + 4b - b + 25 + i\,(12 a -a + 4b) \\ \iff \begin{cases} \begin{align} 2a + 3b + 25 = 0 \\ 11 a + 4b = 0 \end{align} \end{cases} $

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    @TakahiroWaki `It was right that you edit the answer of new user` Sorry, no idea what that's supposed to mean. What I posted was my own answer, a mathematically valid answer, and entirely different from all other posted answers. That said, I am done here and bye.2017-01-12
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If $x^2-4x+5$ is a factor of $ax^3+bx^2+25$, then $ax^3+bx^2+25=(x^2-4x+5)(\text{something})$. Since $ax^3+\cdots$ is a polynomial of degree 3 and $x^2-\cdots$ is a polynomial of degree 2, the "something" must be a polynomial of degree 1: $ax^3+bx^2+0x+25=(x^2-4x+5)(\underline{\;\;\;\;\;\;}x+\underline{\;\;\;\;\;\;})$ Try to fill in the two blanks based the terms on the left side that don't have $a$ and $b$ in them (for example, how will $+25$ end up in the product?), then finish the multiplication to find the values of $a$ and $b$.

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$\rm\ 0 = a\ x^3 + b\ x^2 + 25 - (x^2 - 4\ x + 5)\ (a\ x + 5) = (4\ a + b - 5)\ x^2 + (20 - 5\ a)\ x\ \Rightarrow \ a,\: b = \ldots$

Note: $\rm\ a\ x + 5\ $ comes from comparing leading and constant coefficients.

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$ax^3+bx^2+25=(x^2-4x+5)(ax+k)$ , $k$ = constant

$ax(x^2-4x+5) + k(x^2-4x+5) =0$

$ax^3 + x^2(k-4a)+x(5a-4k)+5k=0$
compare to original then $5k=25 \implies k=5$
and $(5a-4k)=0$ so $a =4$
and $b= k-4a = 5 -16 =-11$