Question

show that (x-2),(x+3) and (x-4) are factors of x3-3x2-10x+24???????????

Answer 1

Answer:

Given polynomial,

p(x) = x³ - 3x² - 10x + 24

To show: ( x - 2 ) , ( x + 3 ) and ( x - 4 ) are factors of p(x)

We use Factor Theorem which states that if x - a is factor of p(x) then p(a) = 0.

we check for ( x - 2 )

p(2) = (2)³ - 3(2)² - 10(2) + 24 = 8 - 12 - 20 + 24 = 32 - 32 = 0

So, ( x - 2 ) is factor of p(x).

we check for ( x - 4 )

p(4) = (4)³ - 3(4)² - 10(4) + 24 = 64 - 48 - 40 + 24 = 88 - 88 = 0

So, ( x - 4 ) is factor of p(x).

we check for ( x + 3 )

p(-3) = (-3)³ - 3(-3)² - 10(-3) + 24 = -27 - 27 + 30 + 24 = 54 - 54 = 0

So, ( x + 3 ) is factor of p(x).

Hence, Proved.

Answer 2

Answer:

Answer:

Given polynomial,

p(x) = x³ - 3x² - 10x + 24

To show: ( x - 2 ) , ( x + 3 ) and ( x - 4 ) are factors of p(x)

We use Factor Theorem which states that if x - a is factor of p(x) then p(a) = 0.

we check for ( x - 2 )

p(2) = (2)³ - 3(2)² - 10(2) + 24 = 8 - 12 - 20 + 24 = 32 - 32 = 0

So, ( x - 2 ) is factor of p(x).

we check for ( x - 4 )

p(4) = (4)³ - 3(4)² - 10(4) + 24 = 64 - 48 - 40 + 24 = 88 - 88 = 0

So, ( x - 4 ) is factor of p(x).

we check for ( x + 3 )

p(-3) = (-3)³ - 3(-3)² - 10(-3) + 24 = -27 - 27 + 30 + 24 = 54 - 54 = 0

So, ( x + 3 ) is factor of p(x).

Hence, Proved.

Step-by-step explanation: