How do you use the remainder theorem to find the remainder of $P (x) = x^{4} - 9 x^{3} - 5 x^{2} - 3 x + 4 \div x + 3$ $P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3$ ?

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How do you use the remainder theorem to find the remainder of $P (x) = x^{4} - 9 x^{3} - 5 x^{2} - 3 x + 4 \div x + 3$ $P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3$ ?

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Answer 1 · 2015-10-23T12:44:15

According to the Remainder Theorem the remainder of a polynomial, $P (x)$ $P(x)$ , divided by $(x + 3)$ $(x+3)$ is $P (- 3)$ $P(-3)$
Remainder $P (- 3) = 292$ $P(-3)= 292$

Explanation:

One method of evaluating $P (- 3)$ $P(-3)$ is to use "synthetic substitution" as demonstrated below:

${: (,"|",1,-9,-5,-3,4), (-3,"|",,-3,36,-93,288), (,,"-----","-----","-----","-----","-----"), (,,1,-12,31,-96,), (,,,,,,R=292) :}$

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How do you use the remainder theorem to find the remainder of $P (x) = x^{4} - 9 x^{3} - 5 x^{2} - 3 x + 4 \div x + 3$ $P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3$ ?

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How do you use the remainder theorem to find the remainder of P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3P(x)=x4−9x3−5x2−3x+4÷x+3P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3?

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Explanation:

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How do you use the remainder theorem to find the remainder of $P (x) = x^{4} - 9 x^{3} - 5 x^{2} - 3 x + 4 \div x + 3$ $P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 div x+3$ ?