Question

in the polynomial xsquare + kx+6what number must be taken as k to get a polynomial for which x-1 is a factor? Find also the other factor of that polynomial
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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• In the polynomial x² + kx + 6, what number must be taken as k to get a polynomial for which x-1 is a factor? Find also the other factor of that polynomial.

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$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given : - \begin{cases} &\sf{polynomial \: f(x) = {x}^{2} + kx + 6} \\ &\sf{x - 1 \: is \: factor \: of \: f(x)} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{the \: value \: of \: k} \\ &\sf{other \: factor \: of \: f(x)} \end{cases}\end{gathered}\end{gathered}$

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Concept used :- Factor Theorem

According to factor theorem,

if f(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number, then, (x-a) is a factor of f(x), if f(a)=0.

Also, we can say,

if (x-a) is a factor of polynomial f(x), then f(a) = 0. 

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$\large\underline\purple{\bold{Solution :- }}$

$\sf \:  ⟼Here, \: f(x) \: = {x}^{2} + kx + 6$

$\sf \:  ⟼Since, \: x - 1 \: is \: factor \: of \: f(x)$

$\bf\implies \:f(1) = 0$

$\sf \:  ⟼ {(1)}^{2} + k \times 1 + 6 = 0$

$\sf \:  ⟼1 + k + 6 = 0$

$\sf \:  ⟼k + 7 = 0$

$\bf\implies \:k = - 7$

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$\bf \:So, \: f(x) = {x}^{2} - 7x + 6$

$\sf \:  ⟼f(x) = {x}^{2} - x - 6x \: + 6$

$\sf \:  ⟼f(x) = x(x - 1) - 6(x - 1)$

$\sf \:  ⟼f(x) = (x - 6)(x - 1)$

$\bf\implies \:other \: factor \: of \: f(x) \: is \: x - 6$

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$\begin{gathered}\begin{gathered}\bf Hence :- \begin{cases} &\sf{value \: of \: k = - 7} \\ &\sf{other \: factor \: of \: f(x) = (x - 6)} \end{cases}\end{gathered}\end{gathered}$

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Answer 2

Answer:

Hope you got your answer