Question

the reminder when 5x^3-x^2+6x-2 is divided by 1-5x is





urgent pls.............,...........

Answer 1

Answer:

Step-by-step explanation:

Hope its helps you:))))))))))

Answer 2

Answer:

The remainder obtained when $5x^3-x^2+6x-2$ is divided by 1-5x =$\frac{-4}{5}$

Step-by-step explanation:

To find,

The remainder when $5x^3-x^2+6x-2$ is divided by 1-5x

Recall the theorem,

Remainder theorem:

When a polynomial p(x) is divided by a linear polynomial (x-a), the remainder obtained =  p(a)

Solution:

Here,

p(x) =  $5x^3-x^2+6x-2$

1-5x = 0 ⇒5x = 1

⇒x = $\frac{1}{5}$

Then, by remainder theorem

The remainder obtained when p(x) =  $5x^3-x^2+6x-2$  is divided by

1-5x is p( $\frac{1}{5}$)

p( $\frac{1}{5}$)  $= 5(\frac{1}{5}) ^3-(\frac{1}{5})^2+6(\frac{1}{5})-2$

$=5(\frac{1}{125}) -(\frac{1}{25})+6(\frac{1}{5})-2\\\\=(\frac{1}{25}) -(\frac{1}{25})+6(\frac{1}{5})-2$

=$\frac{6-10}{5}$

=$\frac{-4}{5}$

The remainder obtained when $5x^3-x^2+6x-2$ is divided by 1-5x =$\frac{-4}{5}$

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