Question

One factor of $x^3-7x+6$ is $x-1$. The other factors are:

• A. $(x-3)(x+2)$
• B. $(x+3)(x-2)$
• C. $(x-3)(x-2)$
• D. $(x+3)(x+2)$

Answer 1

Answer: (B)

$(x+3)(x-2)$

From Factor theorem "The factor theorem states that a polynomial has a factor if and only if $f\left(k\right)=0$
Now given $x^3-7x+6$ is $x-1$
if now from options A If we put values
$=3^3-7\times 3+6=12$
${(-2)}^3-7\times -2+6=12$
Option A does not have factor
Now from option B
$={(-3)}^3-7\times -3+6=0$
$={\left(2\right)}^3-7\times 2+6=0$
From option C
$=3^3-7\times 3+6=12$
$={\left(2\right)}^3-7\times 2+6=0$
From option D
$={(-3)}^3-7\times -3+6=0$
${(-2)}^3-7\times -2+6=12$
Hence the $(x+3)(x-2)$ is the right answer.