zeroes of polynomial;
7y²- 11/3y - 2/3
-----step by step of splitting method anyone??------
Answers
Answer:
The zeroes are 2/3 and -1/7
Step-by-step explanation:
7y^2 - 11y/3 - 2/3
For splitting the middle term, multiply the 1st and the 3rd number.
7y^2 x (-2/3)
= -14y^2/3
Now, find two numbers whose sum will be the 2nd number and product would be similar to the product of 1st and the 3rd number
Here the numbers are -14y/3 and y
Their sum is -14y + 3y/3 = -11/3 (this is the 2nd number)
Their product is -14y/3 x y = -14y^2/3 (this is the product of 1st and the 3rd number)
So, -11/3 should be split into -14y/3 and y
NOTE: For easier splitting, take the number in divide in common, here 1/3, to make the equation 1/3(21y^2 - 11y - 2) and then try to split the middle term. After finding the number, multiply all the numbers with the common number in divide, here 1/3.
7y^2 + y - 14y/3 - 2/3
It is recommended to place the positive numbers with positive ones and the negative numbers with negative ones.
After splitting the middle term, make pairs of numbers and take commons out.
7y^2 + y - 14y/3 - 2/3
= y(7y + 1) - 2/3(7y + 1)
NOTE: Make sure that the numbers in brackets are same or this method would get incorrect. Also see that if there is no number in common, take 1 or -1 as the common number.
After this, use the identity ax +/- bx = (a +/- b)x
Here a = y, b = -2/3 and x = 7y + 1
So the answer would be
y(7y + 1) - 2/3(7y + 1)
= (y - 2/3)(7y + 1)
These are the zeroes as you can equate the numbers to zero, but write one line before that:
On finding zeroes
y - 2/3 = 0 7y + 1 = 0
y = 2/3 7y = -1
y = -1/7
The full answer:
7y^2 - 11y/3 - 2/3
= 7y^2 + y - 14y/3 - 2/3
= y(7y + 1) - 2/3(7y + 1)
= (y - 2/3)(7y + 1)
On finding zeroes
y - 2/3 = 0 7y + 1 = 0
y = 2/3 7y = -1
y = -1/7
Therefore, the zeroes are 2/3 and -1/7
Hope I helped