Question 12/40: Set 1 - Try Out - Mathematics (TKA)

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Given that $(x - 1)$ and $(x + 3)$ are factors of the polynomial equation $x^3 - ax^2 - bx + 12 = 0$

If $x_1, x_2,$ and $x_3$ are the roots of the equation and $x_1 < x_2 < x_3$ , the value of $x_1 + x_2 + x_3$ is ....

Given $P(x) = x^3 - ax^2 - bx + 12$ with factors $(x - 1)$ and $(x + 3)$

If the factor is $x = 1$ , then $P(1) = 0$

(1)^3 - a(1)^2 - b(1) + 12 = 0

-a - b + 13 = 0

a + b = 13 \quad \text{...(i)}

If the factor is $x = -3$ , then $P(-3) = 0$

(-3)^3 - a(-3)^2 - b(-3) + 12 = 0

-27 - 9a + 3b + 12 = 0

-9a + 3b = 15

3a - b = -5 \quad \text{...(ii)}

Eliminate equations $\text{(i)}$ and $\text{(ii)}$

\begin{array}{rcrcrl} 3a & - & b & = & -5 & \\ a & + & b & = & 13 & + \\ \hline 4a & & & = & 8 & \\ a & & & = & 2 & \end{array}

Substitute the value of $a$

a + b = 13

2 + b = 13 \Rightarrow b = 11

Therefore $P(x) = x^3 - 2x^2 - 11x + 12$

Other factors of $P(x)$ are found using Horner with $(x - 1) \Rightarrow x = 1$

\begin{array}{r|rrrrl} & x^3 & x^2 & x^1 & x^0 & \\ \hline 1 & 1 & -2 & -11 & 12 & \\ & & 1 & -1 & -12 & + \\ \hline & 1 & -1 & -12 & 0 & \rightarrow S(x) \\ & x^2 & x^1 & x^0 & & \end{array}

The division result is $S(x) = x^2 - x - 12$

Factor $x^2 - x - 12$

x^2 - x - 12 = (x - 4)(x + 3)

Therefore, the factor besides $(x - 1)$ and $(x + 3)$ is $(x - 4)$

Since $x_1 < x_2 < x_3$ , then

x_1 = -3, \quad x_2 = 1, \quad x_3 = 4

x_1 + x_2 + x_3 = -3 + 1 + 4 = 2

Therefore, the value of $x_1 + x_2 + x_3 = 2$

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