if a,(a+b) and (a+2b) are the zeroes of the polynomial x³-6x²+3x+10,find a and b.
please explain
Answers
Answer:
Given cubic polynomial is
p(x)=x
3
−6x
2
+3x+10
The zeros of the polynomial p(x) are of the form a, a+b and a+2b
Then,
a+a+b+a+2b=−
1
−6
=>3a+3b=6
=>a+b=2 ----------------(i)
Also, a(a+b)+(a+b)(a+2b)+a(a+2b)=
1
3
=>a
2
+ab+a
2
+2ab+ab+2b
2
+a
2
+2ab=3
=>3a
2
+2b
2
+6ab=3 ----(ii)
and a(a+b)(a+2b)=−
1
−10
=>a
3
+a
2
b+2a
2
b+2ab
2
=10
From (i), b=2−a
Putting this value in (ii), we get
=>3a
2
+2(2−a)
2
+6a(2−a)=3
=>3a
2
+2(4−4a+a
2
)+12a−a
2
=3
=>−a
2
+4a+5=0
=>a
2
−4a−5=0
=>a
2
−5a+a−5=0
=>a(a−5)+(a−5)=0
=>.(a−5)(a+1)=0
=>a=5 or a=−1
=>b=−3 or b=3 respectively
From equation (iii), we get
at a=5, b=−3
=>5
3
+(5)
2
(−3)+2(5)
2
(−3)+2(5)(−3)
2
=−10
=>125−75−150+90=−10
=>−10=−10 which is true.
and at a=−1, b=3, we have
=>(−1)
3
+(−1)
2
3+2(−1)
2
(3)+2(−1)(3)
2
=−10
=>−1+3−6−18=10
=>−22=−10 which is not true.
Thus, a=5, b=−3
Zeros of the polynomial are 5, 5−3, 5−2×3 ie 5,2,−1
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