Question 16
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(iii) 5t2+12t+7.
Solution:
To find the zeroes of the polynomial 5t2+12t+7 using the factorization method, we will follow these steps:
Step 1: Write the polynomial
We start with the polynomial:
5t2+12t+7
Step 2: Set the polynomial equal to zero
To find the zeroes, we set the polynomial equal to zero:
5t2+12t+7=0
Step 3: Factor the polynomial
We will use the middle-term splitting method. We need to find two numbers that multiply to 5×7=35 (the product of the coefficient of t2 and the constant term) and add up to 12 (the coefficient of t).
The numbers 5 and 7 satisfy these conditions:
– 5×7=35
– 5+7=12
Now, we can rewrite the middle term 12t using 5t and 7t:
5t2+5t+7t+7=0
Step 4: Group the terms
Next, we group the terms:
(5t2+5t)+(7t+7)=0
Step 5: Factor out the common terms
Now, we factor out the common terms from each group:
5t(t+1)+7(t+1)=0
Step 6: Factor out the common binomial
Now we can factor out the common binomial (t+1):
(5t+7)(t+1)=0
Step 7: Set each factor to zero
Now, we set each factor equal to zero:
1. 5t+7=0
2. t+1=0
Step 8: Solve for t
From the first equation:
5t+7=0⟹5t=−7⟹t=−7/5
From the second equation:
t+1=0⟹t=−1
Step 9: State the zeroes
Thus, the zeroes of the polynomial 5t2+12t+7 are:
t=−7/5 and t=−1
Step 10: Verify the relations between the zeroes and coefficients
Let α=−7/5 and β=−1.
1. Sum of the zeroes:
α+β=−7/5−1=−7/5−5/5=−12/5
According to the relation, −b/a=−12/5 (where b=12 and a=5).
2. Product of the zeroes:
αβ=(−7/5)(−1)=7/5
According to the relation, c/a=7/5 (where c=7 and a=5).
Both relations are verified.
Conclusion
Therefore, the zeroes of the polynomial 5t2+12t+7 are −7/5 and −1.