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Polynomials — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsPolynomials1 Mark
Question
Find the remainder when ${x^3} + 3{x^2} + 3x + 1$ is divided by x + 1
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Answer

x + 1
We need to find the zero of the polynomial x + 1
$x + 1 = 0{\text{ }} \Rightarrow {\text{ }}x = - 1$
While applying the remainder theorem, we need to put the zero of the polynomial x + 1 in the polynomial
${x^3} + 3{x^2} + 3x + 1$,to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( -1 \right) = {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + 3\left( { - 1} \right) + 1$

=-1+3-3+1
= 0
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1$ by x + 1, we will get the remainder as 0.

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