divide the polynomial 2x³-3x²+4x+5÷x+2
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Profile photo for Xain AbidXain Abid
Answered 2 years ago
2x³+3x²-4x-5=0
Its a cubic polynomial and can be solved as
first find factors of the constant in above equation i.e. 5
Factors of 5: ±1,±5
Now use these factors as values of x and out in the given equation to check if these are roots or not. Keep in mind as its cubic equation so we have to find one or three roots.
For x=-1
2x³+3x²-4x-5=0
2(-1)^3+3(-1)^2-4(-1)-5=0
-2+3+4-5=0
1+4-5=0
5-5=0
0=0
So x=-1 is the first root of the given equation
For x=1
2x³+3x²-4x-5=0
2(1)^3+3(1)^2-4*1-5=0
2+3-4-5=0
1-5=0
-4≠0
so x=-1 is not the root of the equation.
Now for x=-5
2x³+3x²-4x-5=0
2(-5)^3+3(-5)^2-4*-5-5=0
-250+75+20-5=0
-160≠0
so x=-5 is also not the root.
so only one roots is found of the given equation as
x=-1
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