Question

the zeroes of the following polynomial?3xsquare+15x+12
1)( both are positive)
2)(both are negative)
3)(postive and one negative)
4)( equal in magnitude but opposite in signs)
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Answer 1

Step-by-step explanation:

3x²+15x+12

3x²+12x+3x+12

3x(x+4)+3(x+4)

(x+4)(3x+3)

x+4 = 0

x = -4

3x+3=0

3x= -3

x = -3/3

x = -1

(2) Both are negative

Answer 2

Given :  3x² + 15x + 12

To find : zeroes of the polynomial :

1)( both are positive)

2)(both are negative)

3)(postive and one negative)

4)( equal in magnitude but opposite in signs)

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Solution:

3x² + 15x + 12 = 0

Divide both sides by 3

=> x² + 5x + 4 = 0

Spilit middle term

=> x² + x + 4x + 4 = 0

=> x(x + 1) + 4(x + 1) = 0

=> (x + 1)(x + 4)  = 0

=> x = - 1 , - 4

Both zeroes are negative  

option 2  (both are negative)

Another method without finding zeros

Product of zeroes = 12/3 = 4

+Ve hence both zeroes has to be either -ve or +ve

Sum of zeroes = - 15/3 = - 5

Sum is -ve

Hence both zeroes has to be - ve  

option 2  (both are negative)

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