Question

if p (x)=x^3-4x^2+x+6, then show that p (3)=0 and hence factorise p (x)

Answer 1

p(x) = x^3 - 4x^2 + x + 6

=> p (3) = ( 3)^3 - 4(3)^2 + (3) + 6

=> p(3) = 27 - 36 + 3 + 6

=> p(3) = 27 - 36 + 9

=> p(3) = 27 + 9 - 36

=> p(3) = 36 - 36

=> p(3) = 0


Now,

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

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

= x^2 ( x - 2) - 2x ( x - 2) - 3(x - 2)

= ( x - 2) (x^2 - 2x - 3)

= (x-2) [ x^2 - 3x + x - 3]

= (x - 2) [ x (x - 3) + 1 (x - 3)]

= (x-2)(x-3) (x+1)

Answer 2

Step-by-step explanation:

p(x) = x^3 - 4x^2 + x + 6

=> p (3) = ( 3)^3 - 4(3)^2 + (3) + 6

=> p(3) = 27 - 36 + 3 + 6

=> p(3) = 27 - 36 + 9

=> p(3) = 27 + 9 - 36

=> p(3) = 36 - 36

=> p(3) = 0

Now,

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

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

= x^2 ( x - 2) - 2x ( x - 2) - 3(x - 2)

= ( x - 2) (x^2 - 2x - 3)

= (x-2) [ x^2 - 3x + x - 3]

= (x - 2) [ x (x - 3) + 1 (x - 3)]

= (x-2)(x-3) (x+1)

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