Question

using the remainder theorem find the remainder when p(x) =x^3-5x^2+3x+1 is divided by x-1/3​

Answer 1

Answer:

simple divde from this ok remember it

Answer 2

$\huge \boxed{ \red{\frac{40}{27} \to1 \frac{13}{27} }}$

Step-by-step explanation:

Tp find -

remainder when p(x) = x³ - 5x² + 3x + 1 divided by x - 1/3 by remainder theorem.

.

Solution -

x - 1/3 = 0

=> x = 1/3

.

Put the value of x in p(x)

$= > {x}^{3} - 5 {x}^{2} + 3x + 1 \\ \\ = > {( \frac{1}{3} )}^{3} - 5( \frac{1}{3} {)}^{2} + 3( \frac{1}{3} ) + 1 \\ \\ = > \frac{1}{27} - \frac{5}{9} + 1 + 1 \\ \\ = > \frac{1 - 15}{27} + 2 \\ \\ = > \frac{ - 14}{27} + 2 \\ \\ = > \frac{ - 14 + 54}{27} \\ \\ = > \frac{40}{27} = 1\frac{13}{27}$

hence remainder is 40/27

hope it helps.

hope it helps.