If one of zeros of cubic …
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Sia ? 6 years, 7 months ago
Let p(x) = x3 + ax2 + bx + c
As -1 is one of the zeroes of p(x), p(-1) = 0
⇒ (-1)3 + a(-1)2 + b(-1) + c = 0
⇒ - 1 + a – b + c = 0
⇒ c = b –a + 1 …. (i)
Let {tex}\alpha{/tex}, β be two other zeroes of p(x),
then product of zeroes =-1× {tex}\alpha{/tex} × β ={tex}\;\;-\frac{\;\;Cons\tan t\;term\;}{coefficient\;of\;x^3}{/tex}
⇒ (-1) (α β) = {tex}\;\;-\frac{\;\;c}1{/tex}
⇒ - {tex}\alpha{/tex} β = - c
⇒ {tex}\alpha{/tex} β = c
⇒ {tex}\alpha{/tex} β = b –a + 1 [using (i)]
Hence, the product of other two zeroes of the given cubic polynomial is b – a + 1.
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