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8x³-ax²-x+2
x = -1/2 is a zero of the polynomial.
Put x = -1/2 in given polynomial.
8(-1/2)³-a(-1/2)²-(-1/2)+2 = 0
8(-1/8) - a(1/4)+1/2+2=0
-1-a/4+1/2+2=0
Take LCM,
LCM is 4
(-4-a+2+8)/4 = 0
-4-a+2+8 = 0
-a+6 = 0
a = 6
Therefore, a = 6
Hope it helps
x = -1/2 is a zero of the polynomial.
Put x = -1/2 in given polynomial.
8(-1/2)³-a(-1/2)²-(-1/2)+2 = 0
8(-1/8) - a(1/4)+1/2+2=0
-1-a/4+1/2+2=0
Take LCM,
LCM is 4
(-4-a+2+8)/4 = 0
-4-a+2+8 = 0
-a+6 = 0
a = 6
Therefore, a = 6
Hope it helps
Anonymous:
I am sorry the question is 8x^3
Answered by
2
Hey there!
f(x)= 8x³-ax²-x+2
Given that -1/2 is a root of f(x).
Which means when x=-1/2, f(x)=0
So substitute x=-1/2 in f(x)
f(-1/2)=8*(-1/2)³-a(-1/2)²-x+2=0
=-8/8-a/4+1/2+2=0
= 2+1/2-1=a/4
=5/2-1=a/4
=5/2-2/2=a/4
=3/2=a/4
a=12/2=6
a=6
f(x)= 8x³-ax²-x+2
Given that -1/2 is a root of f(x).
Which means when x=-1/2, f(x)=0
So substitute x=-1/2 in f(x)
f(-1/2)=8*(-1/2)³-a(-1/2)²-x+2=0
=-8/8-a/4+1/2+2=0
= 2+1/2-1=a/4
=5/2-1=a/4
=5/2-2/2=a/4
=3/2=a/4
a=12/2=6
a=6
sorry the question of the polynomial..In the 1st term it is 8x^3
I've edited my ans
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