[tex] Large{\tt{Question:}} [/tex]
✯ If the sum of the zeroes of the polynomial (a + 1)x² + (2a + 3)x + (3a + 4) is - 1, then find the product of its zeroes.
[tex] Large{\tt{\green{Answer :}}} [/tex]
[tex] {\bigstar{{\boxed { \bf 2}}}}[/tex]
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Answer:
when sum and products of a polynomial are given, then we can form the polynomial as
x^2-x (sum of zeros)+product of zeros....(1)
to bring the given polynomial to the standard form mentioned above( equation 1), we have to divide the given polynomial with the coefficient of x^2 ie by (a + 1) throughout
=> (a+ 1) x^2 - (-1) x (2a + 3) + (3a + 4) ÷ (a + 1)
=>x^2-(-1)x(2a + 3)/(a + 1)+(3a + 4)/(a + 1)......(2)
on comparing equation ( 1 ) and ( 2 )
sum of zeros (coefficient of x) is
-(2a+3)/(a + 1)
and product of zeros is (3a + 4)/(a + 1)
therefore -(2a + 3)/( a + 1) = -1 (given sum of zeros is -1)
=> -2 a - 3 = - a - 1
=> - a = 2
=> a = - 2
therefore product of zeros
= (3a + 4)/(a + 1)
=> (3×- 2 + 4)/(-2 + 1)
=> (-6 + 4)(-2 + 1)
=> (-2)(-1)
=> 2