Question

Divide the polynomial $3x^4-4x^3-3x-1$ by $x-1$. B using long division method only. Also check whether $x+1$ is a factor or not.

Answer 1

$3x^3-x^2-x-4$

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$x-1$$)3x^4-4x^3+0x^2-3x-1$

$3x^4-3x^3$

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$-x^3+0x^2-3x$

$-x^3+x^2$

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$-x^2-3x-1$

$-x^2+x$

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$-4x-1$

$-4x+4$

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$-5$

$\therefore \frac{3x^4-4x^3-3x-1}{x-1}=3x^3-x^2-x-4$

Remainder$=-5$

Now to check whether $x+1$ is factor or not,

$x+1=0$

$x=-1$

Put $x=-1$ in $3x^4-4x^3-3x-1$

$3(-1{)}^4-4(-1{)}^3-3(-1)-1$

$=3+4+3-1$

$=9\ne 0$

$\therefore$ By Remainder theorem, we conclude $(x+1)$ is not a factor.