Question

p(x) : 5x³- x² + 6x - 2 ; 1 - 5x
solve through remainder theorem.
​

Answer 1

Step-by-step explanation:

Given :-

P(x) = 5x³- x² + 6x - 2

Divisor = 1 - 5x

To find :-

Solve through remainder theorem.?

Solution :-

Given that

P(x) = 5x³- x² + 6x - 2

Divisor= 1 - 5x

We know that

Remainder Theorem:

Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divided by x-a then the remainder is P(a).

Now

If P(x) is divided by 1-5x then the remainder is P(1/5).

Since 1-5x = 0 => 1 = 5x => x = 1/5

Now,

=> P(1/5) = 5(1/5)³-(1/5)²+6(1/5)-2

=> P(1/5) = 5(1/125)-(1/25)+(6/5)-2

=> P(1/5) = (5/125)-(1/25)+(6/5)-2

=> P(1/5) = (1/25)-(1/25)+(6/5)-2

=> P(1/5) = [(1-1)/25]+(6/5)-2

=> P(1/5) = (0/25)+(6/5)-2

=> P(1/5) = 0+(6/5)-2

=> P(1/5) = (6/5)-2

=> P(1/5) = (6-10)/5

=> P(1/5) = -4/5

Therefore, P(1/5) = -4/5

Answer:-

The remainder for the given problem is -4/5

Check:-

Divisor = 1-5x = -5x+1

On dividing p(x) by -5x+1 then

-5x+1) 5x³- x² + 6x - 2(-x²-(6/5)

5x³ -x²

(-) (+)

______________

6x-2

6x -(6/5)

(-) (+)

_______________

(6/5)-2

________________

Quotient = -x²-(6/5)

=> Remainder = (6/5)-2

=> Remainder = (6-10)/5

=> Remainder = -4/5

Verified the given relations in the given problem.

Used formulae:-

Remainder Theorem:-

"Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divided by x-a then the remainder is P(a)".