If x^3-5x^2-px+24=(x-4).q (x). what is the value of p
Answers
Answered by
21
Substitute x = 4
4³ - 5(4²) - 4p + 24 = 0
64 - 80 - 4p + 24 = 0
p = 2
4³ - 5(4²) - 4p + 24 = 0
64 - 80 - 4p + 24 = 0
p = 2
VinilM:
but how 4?
in the question it is given right
to make rhs zero, substitute x = 4
ok thanks got it.
hmm
Answered by
18
Hello Dear!!!
Here's your answer...
Let,
f(x) = x^3 - 5x^2 - px +24 = 0
Zero of x- 4 is
x - 4 = 0
x = 4
Substitute the value of x
x^3 - 5x^2 - px +24 = 0
(4)^3 - 5(4)^2 - p(4) + 24 = 0
64 - 80 - 4p + 24 = 0
8 - 4p = 0
4p = 8
p = 2
The value of p is 2
____________________________
Hope this helps you....
Here's your answer...
Let,
f(x) = x^3 - 5x^2 - px +24 = 0
Zero of x- 4 is
x - 4 = 0
x = 4
Substitute the value of x
x^3 - 5x^2 - px +24 = 0
(4)^3 - 5(4)^2 - p(4) + 24 = 0
64 - 80 - 4p + 24 = 0
8 - 4p = 0
4p = 8
p = 2
The value of p is 2
____________________________
Hope this helps you....
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