Question 23
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(x) 7y2−(11/3)y−2/3
Solution:
To find the zeroes of the polynomial 7y2−(11/3)y−2/3 using the factorization method, we will follow these steps:
Step 1: Write the polynomial equation
We start with the given polynomial:
7y2−(11/3)y−2/3=0
Step 2: Eliminate fractions
To simplify the calculations, we can eliminate the fractions by multiplying the entire equation by 3:
3(7y2)−3[(11/3)y]−3(2/3)=0
This simplifies to:
21y2−11y−2=0
Step 3: Factor the quadratic
Next, we need to factor the quadratic equation 21y2−11y−2. We look for two numbers that multiply to 21×(−2)=−42 and add up to −11.
The numbers that satisfy these conditions are −14 and 3.
Step 4: Rewrite the middle term
We can rewrite the equation using these numbers:
21y2−14y+3y−2=0
Step 5: Group the terms
Now, we group the terms:
(21y2−14y)+(3y−2)=0
Step 6: Factor by grouping
Factoring out the common factors from each group gives us:
7y(3y−2)+1(3y−2)=0
Now we can factor out (3y−2):
(3y−2)(7y+1)=0
Step 7: Find the zeroes
Setting each factor to zero gives us the zeroes:
1. 3y−2=0 → y=2/3
2. 7y+1=0 → y=−1/7
Thus, the zeroes of the polynomial are:
y=2/3 and y=−1/7
Step 8: Verify the relations between zeroes and coefficients
Let α=2/3 and β=−1/7.
Sum of the zeroes:
α+β=2/3−1/7
To add these fractions, we find a common denominator (21):
α+β=14/21−3/21=11/21
According to the relation, the sum of the zeroes is given by −ba:
−(−11/21)=11/21
Product of the zeroes:
αβ=(2/3)(−1/7)=−2/21
According to the relation, the product of the zeroes is given by c/a:
−2/21
Both relations hold true, confirming that our factorization and calculations are correct.