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Question
The factors of x3−7x+6 are
The factors of x3−7x+6 are
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Solution
The correct option is A (x-1), (x-2), (x+3)
Let's take p(x) = x3−7x+6By trail and error method,p(1) = 13−7(1)+6=1−7+6=0so x−1 is a factor1
\begin{tabular}{|llll}
1 & 0 & \( -7 \) & 6 \\
0 & 1 & 1 & \( -6 \) \\
\hline 1 & 1 & \( -6 \) & 0
\end{tabular}x2+x−6 =x2+3x−2x−6=x(x+3)−2(x+3)The factors are (x−1)(x−2) and (x+3)Hence Option B is the correct answer.
Let's take p(x) = x3−7x+6By trail and error method,p(1) = 13−7(1)+6=1−7+6=0so x−1 is a factor1 \begin{tabular}{|llll} 1 & 0 & \( -7 \) & 6 \\ 0 & 1 & 1 & \( -6 \) \\ \hline 1 & 1 & \( -6 \) & 0 \end{tabular}x2+x−6 =x2+3x−2x−6=x(x+3)−2(x+3)The factors are (x−1)(x−2) and (x+3)Hence Option B is the correct answer.
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