Question

Factorise by Remainder Theorem:- x³-3x²-9x-5​

Answer 1

Answer:

The answer is (x-1)(x-1)(x-5).

Step-by-step explanation:

Question:x^3-3x^2-9x-5

Here constant term=-5

Factors of -5=+_1,+_2,+_3,+_4,+_5

Let p(X)=x^3-3x^2-9x-5

Put X=5

Remainder=p(5)

=(5)^3-3×(5)^2-9(5)-5

=125-125

=0

Therefore,(x-5) is a factor of p(X)

Now ,p(X)=(X)^3-3×(X)^2-9x-5

g(X)=(x-5)

When we divide it ,we will get the quotient

i.e.(X)^2+2x+1

p(X)=quotient×divisor +remainder

=((X)^2+2x+1)(x-5)+0

= we will get

=(x-1)(x-1)(x-5).

Hope it will help u.

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Answer 2

X³ - 3x² - 9x - 5

= x³ + x² - 4x² - 4x  - 5x - 5

= x²(x+1) - 4x(x+1) -5(x+1)

= (x+1)(x² - 4x - 5)

= (x+1)(x² - 5x + x - 5)

= (x+1)[x(x-5) + 1(x-5)]

= (x+1)[(x-5)(x+1)]

= (x+1)(x+1)(x-5)

=(x+1)²(x-5)