Question

x+1 is a factor of the polynomial i) x^3+x^2-x+1 ii) x^3+x^2+x+1 iii) x^4+x^3+x^2+1 iv) x^4+3x^3+3x^2+x+1​

Answer 1

This is a simple question.

Just substitute the value of x that is x+1=0 -> x=-1

and u will get the answer.

I hope it helps

Answer 2

$\bold{\tt \underline{ \underline{Concept \: of \: Remainder \: theorem : }} }$

$\sf : \implies If \: p(x) \: is \: divided \: by \: x+a, \: then \: the \: remainder \: is \: p(-a)$

$\sf : \implies If \: p(x) \: is \: divided \: by \: x - a, \: then \: the \: remainder \: is \: p(a)$

$\sf : \implies Similarly, \: if \: f(x) \: is \: divided \: by \: 2x-1, \: then \: the \: remainder \: is \: p\bigg( \dfrac{1}{2} \bigg)$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\sf First \: find \: the \: zero \: of \: x+1, \: which \: is \: \bf \underline{x+1=0} \sf \: \: or \: \underline{ \: \bf x=-1}$

$\bf \underline{Now},$

$\bf (i) \: \sf \: x^{3} + x^{2} - x + 1$

$\sf \implies p( - 1) = ( - 1)^{3} + ( - 1)^{2} -( - 1 )+ 1$

$\sf \implies- 1 + 1 + 2 = 2$

$\sf This \: is \: not \: zero,$

$\sf Then \: x+1 \: is \: not \: the \: factor \: of \: equation.$

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$\bf (ii) \: \sf \: x^{3}+x^{2}+x+1$

$\sf \implies p( - 1) = ( - 1)^{3}+( - 1)^{2}+( - 1)+1$

$\sf \implies - 1+1+0 = 0$

$\sf Then \: x+1 \: is \: factor \: of \: equation.$

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$\bf (iii) \: \sf \: x^4+x^3+x^2+1$

$\sf \implies p( - 1) = \: ( - 1)^4+( - 1)^3+( - 1)^2+1$

$\sf \implies 1+( - 1)+1+1$

$\sf \implies 0+2 = 2$

$\sf This \: is \: not \: zero,$

$\sf Then \: x+1 \: is \: not \: the \: factor \: of \: equation.$

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$\bf (iv) \: \sf \: x^4+3x^3+3x^2+x+1$

$\sf \implies p( - 1) = \:( - 1)^4+3( - 1)^3+3( - 1)^2+( - 1)+1$

$\sf \implies 1+( - 3)+3+0$

$\sf \implies - 2+3 = 1$

$\sf This \: is \: not \: zero,$

$\sf Then \: x+1 \: is \: not \: the \: factor \: of \: equation.$