Factor the polynomial f(x) then solve f(x)=0?

Question

A. F(x)=x^3-6x^2+3x+10 

B. F(x)=x^3-x^2-14x+24

Answer 1

A).

F( x ) = x^3 - 6x^2 + 3x + 10.

Factor x^3 - 6x^2 + 3x + 10 by factor by grouping.

= x^3 - 5x^2 - x^2 + 3x + 10

= x^2(x - 5) - (x^2 - 3x - 10)

= x^2(x - 5) - (x^2 - 5x + 2x - 10)

= x^2(x - 5) - [ x(x - 5) + 2(x - 5 ]

= x^2(x - 5) - (x - 5)(x + 2)

= (x - 5)[ x^2 - x - 2]

= (x - 5)[ x^2 - 2x + x - 2]

= (x - 5)[ x(x - 2) + 1( x - 2) ]

= (x - 5)[ (x - 2) (x + 1) ]

F( x ) = (x - 5)(x - 2)(x + 1).

Let, F( x ) = (x - 5)(x - 2)(x + 1) = 0

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

Apply zero product property.

x - 5 = 0, x - 2 = 0, and x + 1 = 0

⇒ x = 5, x = 2, and x = - 1.

Solution of F( x ) is x = 5, x = 2, and x = - 1.

B).

F( x ) = x^3 - x^2 - 14x + 24.

Factor x^3 - x^2 - 14x + 24 by factor by grouping.

= x^3 - 2x^2 + x^2 - 14x + 24

= x^2(x - 2) + (x^2 - 14x + 24)

= x^2(x - 2) + (x^2 - 12x - 2x + 24)

= x^2(x - 2) + [ x(x - 12) - 2(x - 12) ]

= x^2(x - 2) + [ (x - 12)(x - 2) ]

= (x - 2)[ x^2 + x - 12 ]

= (x - 2)[ x^2 + 4x - 3x - 12 ]

= (x - 2)[ x(x + 4) - 3(x + 4) ]

= (x - 2)(x + 4)(x - 3)

F( x ) = (x - 2)(x + 4)(x - 3).

Let, F( x ) = (x - 2)(x + 4)(x - 3) = 0

(x - 2)(x + 4)(x - 3) = 0

Apply zero product property.

x - 2 = 0, x + 4 = 0, and x - 3 = 0

⇒ x = 2, x = - 4, and x = 3.

Solution of F( x ) is x = 2, x = - 4, and x = 3.