Question

find the remainder when 3 x cube minus 6 X square + 3 x minus 7 by 9 is divided by 3 x minus 4



Step by Step By P(x) method​

Answer 1

$\textbf{Concept:}$

$\text{Remainder theorem:}$

$\text{The remainer when P(x) is divided by (ax-b) is P(\frac{b}{a}) }$

$\textbf{Given:}$

$\text{P(x)=}3x^3-6x^2+3x-7$

$\text{The remainer when P(x) is divided by (3x-4) is P(\frac{4}{3}) }$

$P(\frac{4}{3})$

$=3(\frac{4}{3})^3-6(\frac{4}{3})^2+3(\frac{4}{3})-7$

$=3(\frac{64}{27})-6(\frac{16}{9})+3(4)-7$

$=\frac{64}{9}-2(\frac{16}{3})+5$

$=\frac{64}{9}-\frac{32}{3}+5$

$=\frac{64-96+45}{9}$

$=\frac{13}{9}$

$\therefore\textbf{The remainer when P(x) is divided by (3x-4) is \bf\frac{13}{9} }$

Find more:

If a quadratic equation of the form ax^2 + c when divided by x and (x + 1) leaves remainder 2 and 4 respectively, then the value of a^2 + c^2 is

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Answer 2

Answer:

Hii

Step-by-step explanation:

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