3) show that (x-2). (x+3) and (x-4) are the
factors of x-3x² - 10x+ 24
Answers
Answer:
Given, (x−2) , (x+3) and (x−4) are factors of polynomial x
3
−3x
2
−10x+24.
Then, f(x)=x
3
−3x
2
−10x+24.
If (x−2) is a factor, then x−2=0⟹x=2.
Replace x by 2, we get,
f(2)=(2)
3
−3(2)
2
−10(2)+24
f(2)=8−12−20+24
f(2)=0.
The value of f(2) is zero.
Then (x−2) is the factor of the polynomial x
3
−3x
2
−10x+24.
If (x+3) is factor, then x+3=0⟹x=−3.
Replace x by −3, we get,
f(−3)=(−3)
3
−3(−3)
2
−10(−3)+24
f(−3)=−27−27+30+24
f(−3)=0.
The value of f(−3) is zero.
Then (x+3) is the factor of the polynomial x
3
−3x
2
−10x+24.
If (x−4) is factor, then x−4=0⟹x=4.
Replace x by 4, we get,
f(4)=(4)
3
−3(4)
2
−10(4)+24
f(4)=64−48−40+24
f(4)=0.
The value of f(4) is zero.
Then (x+4) is the factor of the polynomial x
3
−3x
2
−10x+24.
Therefore, hence showed that (x−2) , (x+3) and (x−4) are factors of polynomial x
3
−3x
2
−10x+24.
Answer:
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Step-by-step explanation:
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).