Question 22
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(ix) y2+(3√5/2)y−5
Solution:
To find the zeroes of the polynomial y2+(3√5/2)y−5 using the factorization method, we will follow these steps:
Step 1: Write the polynomial in standard form
We start with the polynomial:
y2+(3√5/2)y−5=0
Step 2: Identify coefficients
Here, we identify the coefficients:
– a=1 (coefficient of y2)
– b=3√5/2 (coefficient of y)
– c=−5 (constant term)
Step 3: Use the middle-term splitting method
We need to express the middle term 3√5y/2 in a way that allows us to factor the polynomial. We look for two numbers that multiply to ac=1×(−5)=−5 and add up to b=3√5/2.
The two numbers that satisfy these conditions are 2√5 and −5/2.
Step 4: Rewrite the polynomial
We can rewrite the polynomial as:
y2+2√5y−5y/2−5=0
Step 5: Factor by grouping
Now we group the terms:
(y2+2√5y)+(−5y/2−5)=0
Factoring out common terms:
y(y+2√5)−5/2(y+2√5)=0
Now we can factor out (y+2√5):
(y−5/2)(y+2√5)=0
Step 6: Find the zeroes
Setting each factor to zero gives us:
1. y−5/2=0 → y=5/2
2. y+2√5=0 → y=−2√5
Thus, the zeroes of the polynomial are:
y=5/2 and y=−2√5
Step 7: Verify the relations between the zeroes and coefficients
Let α=5/2 and β=−2√5.
1. Sum of the zeroes:
α+β=5/2−2√5
We need to check if this equals −b/a:
−b/a=−(3√5/2)/1=−3√5/2
2. Product of the zeroes:
αβ=(5/2)(−2√5)=−5√5
We need to check if this equals ca:
c/a=−5/1=−5
Conclusion
The zeroes of the polynomial y2+(3√5/2)y−5 are 5/2 and −2√5. The relations between the zeroes and coefficients are verified.