$p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2$

daniko883y
2022-09-04
Answered

Check whether p(x) is a multiple of g(x) or not

$p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2$

$p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2$

You can still ask an expert for help

Elisa Spears

Answered 2022-09-05
Author has **9** answers

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Step 1

Given

$p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2Step\; 2\; Finding\; the\; given$

According to the question,

$=g(x)=x-2$

$g(x)=0$

$x-2=0$

$x=2$

zero of $g(x)=2$

So, substituning the value of x in p(x).

$p(2)=(2{)}^{3}-5(2{)}^{2}+4(2)-3$

$p(2)=(2{)}^{3}-5(2{)}^{2}+4(2)-3$

$=-7\ne 0$

\therefore p(x) is not the multiple of g(x) since the remainder $\ne 0$

Given

$p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2Step\; 2\; Finding\; the\; given$

According to the question,

$=g(x)=x-2$

$g(x)=0$

$x-2=0$

$x=2$

zero of $g(x)=2$

So, substituning the value of x in p(x).

$p(2)=(2{)}^{3}-5(2{)}^{2}+4(2)-3$

$p(2)=(2{)}^{3}-5(2{)}^{2}+4(2)-3$

$=-7\ne 0$

\therefore p(x) is not the multiple of g(x) since the remainder $\ne 0$

Inbrunstlr

Answered 2022-09-06
Author has **2** answers

Given, $p(x)={x}^{3}-5{x}^{2}+4x-3$

$g(x)=x-2$

$\Rightarrow x-2=0$

$\Rightarrow x=2$

Now, p(2) should be 0 if p(x) is a multiple of g(x)

Thus,

$p(2)={2}^{3}-5(2{)}^{2}+4(2)-3$

$=8-20+8-3$

$=-7$

$=8-20+8-3$

$=0$

Thus, p(x) is not a multiple of g(x)

$g(x)=x-2$

$\Rightarrow x-2=0$

$\Rightarrow x=2$

Now, p(2) should be 0 if p(x) is a multiple of g(x)

Thus,

$p(2)={2}^{3}-5(2{)}^{2}+4(2)-3$

$=8-20+8-3$

$=-7$

$=8-20+8-3$

$=0$

Thus, p(x) is not a multiple of g(x)

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