find the value of and b so that the polynomial x cube-ax square -13 x+b has x-1 and x+3 as factors.
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f(x) = x³ -ax² -13x + b
A/Q (x-1), (x+3) are factors of f(x)
Implies 1, -3 are zeroes of f(x)
S o f(1) = (1)³ - a(1)² - 13(1) + b = 0 ⇒ 1-a -13 + b = 0 ⇒ b-a = 12 ...(1)
And f(-3) = (-3) ³ -a(-3)² -13(-3) + b =0⇒ -27 -9a +39 + b =0
⇒ b-9a = -12 ........(2)
Solving fetches a = -3 and b = 9
Hope my answer is correct.
A/Q (x-1), (x+3) are factors of f(x)
Implies 1, -3 are zeroes of f(x)
S o f(1) = (1)³ - a(1)² - 13(1) + b = 0 ⇒ 1-a -13 + b = 0 ⇒ b-a = 12 ...(1)
And f(-3) = (-3) ³ -a(-3)² -13(-3) + b =0⇒ -27 -9a +39 + b =0
⇒ b-9a = -12 ........(2)
Solving fetches a = -3 and b = 9
Hope my answer is correct.
parisakura98pari:
Yes its 3x " - 2" , you told that earlier and putting x = 2 , it produces 2(2)^4 -6(2)^3 +3(2)^2 + 3(2) - 2 = 32 - 48 +12 +6-2 = 0
This proves the question.
when we put one it is coming 4
Nope, you check .
Again reminding you it's -2 , given in question>
ok i understood thanks very much
Really, don't hesitate to ask more.
no i have only 2 you have solved thaks i will ask next time
byee
hi i am back
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