x2 + 11 x + 30 please answer this question in 1 hour
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Heya !!!
This is the answer of your question....
Given polynomial :---
x² + 11x + 30.
We have to find two numbers a and b such that a+b =11 and ab =30.
By looking at prime factors of 30,
30 = 2×3×5.
So, we can say that a and b are 6 and 5.
x² + 11x + 30 =0
x² + 6x + 5x + 30 = 0
x (x +6) + 5(x +6) = 0
(x+5)(x+6) =0
(x+5) = 0 or. (x+6)=0
x = -5 or x = -6.
Hence, the two zeroes of given polynomial are -5 and -6.
Hope you got the right answer...
This is the answer of your question....
Given polynomial :---
x² + 11x + 30.
We have to find two numbers a and b such that a+b =11 and ab =30.
By looking at prime factors of 30,
30 = 2×3×5.
So, we can say that a and b are 6 and 5.
x² + 11x + 30 =0
x² + 6x + 5x + 30 = 0
x (x +6) + 5(x +6) = 0
(x+5)(x+6) =0
(x+5) = 0 or. (x+6)=0
x = -5 or x = -6.
Hence, the two zeroes of given polynomial are -5 and -6.
Hope you got the right answer...
Answered by
0
Heya,
x^2 + 11x + 30 [To find the zeros of the given polynomial]
=> Splitting the middle term
[Product = 30 ; Sum = 11]
=> x^2 + 6x + 5x + 30
=> x (x + 6) 5(x + 6)
=> (x + 5) (x + 6) are the factors of the x^2+11x+30
=> Zeros of polynomial is -5 & -6
Hope my answer helps you :)
Regards,
Shobana
x^2 + 11x + 30 [To find the zeros of the given polynomial]
=> Splitting the middle term
[Product = 30 ; Sum = 11]
=> x^2 + 6x + 5x + 30
=> x (x + 6) 5(x + 6)
=> (x + 5) (x + 6) are the factors of the x^2+11x+30
=> Zeros of polynomial is -5 & -6
Hope my answer helps you :)
Regards,
Shobana
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