What are the possible expressions for the dimensions of the cuboid whose volume=6y^3+y^2-12y+5
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Given volume of cuboid = 6y^3+y^2-12y+5 .... (1)
In order to find its dimensions we have to factorize the given expression as:
On putting the y = 1 in (1), we get
f (1) = 6 × (1)³+ (1)² - 12 × (1) + 5
= 6 + 1 - 12 +5
= 0.
So y-1 is factor of polynomial p(y).
(6y³+y²-12y+5)/y-1
= 6y²+7y-5
= 6y²+10y-3y-5
= 2y(3y+5) - 1(3y+5)
= (2y-1) (3y+5)
Hence the expressions for the dimensions of the cuboid are (y-1), (2y-1) and (3y+5), whose volume is 6y^3+y^2-12y+5.
In order to find its dimensions we have to factorize the given expression as:
On putting the y = 1 in (1), we get
f (1) = 6 × (1)³+ (1)² - 12 × (1) + 5
= 6 + 1 - 12 +5
= 0.
So y-1 is factor of polynomial p(y).
(6y³+y²-12y+5)/y-1
= 6y²+7y-5
= 6y²+10y-3y-5
= 2y(3y+5) - 1(3y+5)
= (2y-1) (3y+5)
Hence the expressions for the dimensions of the cuboid are (y-1), (2y-1) and (3y+5), whose volume is 6y^3+y^2-12y+5.
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