Question

If (a - 1) is a factor of a³ - ka² + 11a - 6, then the value of k should be:​

Answer 1

$\sf{If (a - 1) is \: the \: factor \: of \: a^3 - ka^2 +11a -6 }$

$\textsf{Then, the value of a will be :}$

$\sf\longmapsto a - 1 = 0 \\ \sf \green{\longmapsto a = 1}$

Now, putting the value of a in the given

$\sf\implies p(a) = a³ - ka² + 11a - 6$

$\sf\implies p(1) = 1³ - k(1)² + 11(1) - 6$

$\sf\implies p(1) = 1 - k + 11 - 6$

$\sf\implies p(1) = 12 - k$

Now, as (a - 1) is the factor/ root of the given polyⁿ.So, we can equate the whole eqⁿto 0.

$\displaystyle{\therefore}$

$\sf{ \looparrowright 12 - k = 0 }$

$\sf \large \purple{ \looparrowright k = 12}$