Question

answer this and will mark as brainlist..these 2 questions.. pls

Answer 1

p(x) = x cube - 4x sq + x + 6

Given p( 3 ) = 0

p(3) = 3 cube - 4(3 sq) + 3 + 6

p(3) = 27 - 4(9) + 9

p(3) = 27 - 36 + 9

p(3) = 36 - 36

p(3) = 0

Therefore it is proved that p(3) = 0

Factorisation

x^3 - 4x^2 + x + 6

(x^3 - 2x^2) - (2x^2 + x + 6)

x^(x - 2) - (x - 2)(2x + 3)

Common factor is ( x - 2 ) and other is
( x^2 - 2x - 3 )

Now factorise ( x^2 - 2x - 3 )

x^2 - 3x + 1x - 3

x(x - 3) + 1(x - 3)

For this the common factor is (x - 3) and the other is (x + 1)

Therefore the factored form of......

x^3 - 4x^2 + x + 6 is (x-2) , (x-3) , (x+1)


Hope it helps u.....