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$\text{Đa thức bậc nhất có dạng: ${f(x)}$ = ax + b}$

$\text{Ta có:}$

${f(-2)}$ $\text{= -11}$

$\Rightarrow$ $\text{a(-2) + b = -11}$

$\Rightarrow$ $\text{-2a + b = -11 ( 1 )}$

${f(4)}$ $\text{= -10}$

$\Rightarrow$ $\text{a.4 + b = - 10}$

$\Rightarrow$ $\text{4a + b = - 10 ( 2 )}$

$\text{Từ ( 1 ) và ( 2 ), ta có hệ phương trình:}$

$\begin{cases} -2a + b = -11 ( 1 )\\4a + b = - 10 ( 2 ) \end{cases}$

$\begin{cases} -2a + b = -11 ( 1 )\\-\\4a + b = - 10 ( 2 ) \end{cases}$

$\text{-6a = -1}$

$\text{a =}$ $\dfrac{1}{6}$

$\text{Thay a =}$ $\dfrac{1}{6}$ vào ( 2 ), ta được:

$\text{4.$\dfrac{1}{6}$ + b = - 10}$

$\text{b =}$ $\dfrac{-32}{3}$

$\text{Vậy đa thức cần tìm là:}$ ${f(x)}$ = $\dfrac{1}{6}$x - $\dfrac{32}{3}$