Day: October 25, 2021

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202110251056 Solution to 1970-CE-AMATH-1

Given that F(x)=(b-c)x^3-(a+2b-3c)x^2-(b+2c-3a)x-2(a-b). By using the remainder theorem, prove that (x-1)(x-2) is a factor of F(x). Find also the remaining factor of F(x).

Solution.

\begin{aligned} F(1) & = (b-c)(1)^3 - (a+2b-3c)(1)^2 - (b+2c-3a)(1) - 2(a-b) = 0 \\ F(2) & = (b-c)(2)^3 - (a+2b-3c)(2)^2 - (b+2c-3a)(2) - 2(a-b) = 0 \\ \end{aligned}

\therefore (x-1)(x-2) is a factor of F(x).

By trial-and-error,

\begin{aligned} F(3) & = (b-c)(3)^3-(a+2b-3c)(3)^2-(b+2c-3a)(3)-2(a-b) \\ & = -2a+8b-6c \\ & \neq 0 \\ F(4) & = (b-c)(4)^3-(a+2b-3c)(4)^2-(b+2c-3a)(4)-2(a-b) \\ & = -6a+30b-24c \\ & \neq 0 \\ F(5) & = (b-c)(5)^3-(a+2b-3c)(5)^2-(b+2c-3a)(5)-2(a-b) \\ & = -12a+112b-60c \\ & \neq 0 \\ F(6) & = (b-c)(6)^3-(a+2b-3c)(6)^2-(b+2c-3a)(6)-2(a-b) \\ & = -20a+140b-120c \\ & \neq 0 \\ F(7) & = (b-c)(7)^3-(a+2b-3c)(7)^2-(b+2c-3a)(7)-2(a-b) \\ & = -30a+240b-210c \\ & \neq 0 \\ F(8) & = (b-c)(8)^3-(a+2b-3c)(8)^2-(b+2c-3a)(8)-2(a-b) \\ & = -42a+378b-336c \\ & \neq 0 \\ F(9) & = (b-c)(9)^3-(a+2b-3c)(9)^2-(b+2c-3a)(9)-2(a-b) \\ & = -56a + 560b - 504c \\ & \neq 0 \end{aligned}

It may be feasible to work in this manner, however endlessly, to find a zero. But it is high time for me to work in another way round.

By long division, the remaining factor is

(b-c)x+(b-a).

One should check that

\begin{aligned} &\quad (x-1)(x-2)\big( (b-c)x+(b-a)\big) \\ & = (x^2-3x+2)\big( (b-c)x+(b-a)\big) \\ & = (b-c)x^3 - (a+2b-3c)x^2 - (b+2c-3a)x - 2(a-b) \\ & = F(x) \\ \end{aligned}

Remark.

This remaining zero, i.e., \displaystyle{x=\frac{b-a}{b-c}}, is hard to find by trial-and-error, so long division is necessary.

Roughwork.

\begin{aligned} & (b-c)x & +(-a+b) & & \\\cline{2-5} x^2-3x+2\quad \Big)& (b-c)x^3 & +(-a-2b+3c)x^2 & +(3a-b-2c)x & +(-2a+2b) \\ & (b-c)x^3 & +(-3b+3c)x^2 & + (2b-2c)x & \\\cline{2-5} & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\ & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\\cline{3-5} \end{aligned}