What must be added to the sum of 2a2−3a+7,−5a2−2a−11 and 3a2+5a−8 to get 0?
12
−12
a2+a
a−1
A
a2+a
B
−12
C
a−1
D
12
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Solution
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Let 'x' be added to these polynomial to get 0. ⇒(2a2−3a+7)+(−5a2−2a−11)+(3a2+5a−8)+x=0 ⇒(2a2−5a2+3a2)+(−3a−2a+5a)+(7−11−8)+x=0 ⇒0+0+(−12)+x=0 ⇒x=12 Option B is correct.
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