Two zeroes of a cubic polynomial …
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Sia ? 5 years, 11 months ago
Let p(x)=ax3+3x2−bx−6
∵ - 1 is a zero
∴ p(−1)=0
⇒ a(−1)3+3(−1)2−b(−1)−6=0
⇒ −a+3+b−6=0
⇒ −a+b=3..........(i)
Also, -2 is another zero
∴ p(−2)=0
⇒ a(−2)3+3(−2)2−b(−2)−6=0
⇒ −8a+12+2b−6=0
⇒ −8a+2b=−6
⇒ 4a−b=3.........(ii)
Solving equation (i) and (ii), we get
a = 2 and b = 5
∴ p(x) = 2x3 + 3x2 - 5x - 6
Let third zero = k
Sum of zeroes = −32
⇒ -1 + (-2) + k = −32⇒k=−32+3=32
Therefore, third zero = 32
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