how did that step occur. that -1 is also root.

how did that step occur. that -1 is also root. FRiENDi S posted null Dear student " 7:20 PM Let the four terms be a, ar,ap and a-pap = 112 ar + ap 48 Dividing eq (1) by eq (2), we get a+ar3 ar+ar2 112 48 3 3 + 3r3 7r + 7r2 3r3 — 7r2 — 7r+3=O , -1 is a root . So, (r + 1) is a Also Now, (r + 1) (3r2 -IOr+3) = o -9r-r+3) = O (r + 1) (3r(r - 3) - 1 (r - 3)) = O Rut r = -1 (not noggihlp as 1 + r

Dear student,
Here is the explanation of your doubts:
Here we have;3r3-7r2-7r+3=0To find the roots here we apply hit and trial rule.By using hit and trial rule if we put, r=-1 we have;3-13-7-12-7-1+3=-3-7+7+3 = 0This means -1 is root of this cubic polynomial.
Hope it clears your doubt! If you have still any doubts kindly post the solution along with the complete query.

Regards

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