Question

by using long division method show that x minus 3 is a factor of 6 + X - 4 x square + x cube​

Answer 1

$(x^3-4x^2+x+6)\div\;x-3$

$\begin{array}{r|l}&x^2-x-2\\\cline{2-2}\\x-3&x^3-4x^2+x+6\\&x^3-3x^2\\\cline{2-2}&\;\;\;\;\;-x^2+x\\&\;\;\;\;\;-x^2+3x\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;-2x+6\\&\;\;\;\;\;\;\;\;\;\;\;\;-2x+6\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\\\cline{2-2}\end{array}$

$\text{From the above long division method, it is clear that }$

$\text{when x^3-4x^2+x+6 is divided by x-3 the remainder is 0}$

$\implies\text{x-3 is a factor}$

Answer 2

Step-by-step explanation:

We need to show that (x-3) is a factor of $(6+x-4x^2+x^3)$ using long division method. We can write it as $(x^3-4x^2+x+6)$

$\begin{array}{r|l}&x^2-x-2\\\cline{2-2}\\x-3&x^3-4x^2+x+6\\&x^3-3x^2\\\cline{2-2}&\;\;\;\;\;-x^2+x\\&\;\;\;\;\;-x^2+3x\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;-2x+6\\&\;\;\;\;\;\;\;\;\;\;\;\;-2x+6\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\\\cline{2-2}\end{array}$

The above long division method shows that the remainder at the end is equal to 0. It would mean that (x-3) is a factor of $(x^3-4x^2+x+6)$.

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