Question 15
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(ii) 3x2+4x−4
Solution:
To find the zeroes of the polynomial 3x2+4x−4 using the factorization method, we will follow these steps:
Step 1: Set the polynomial equal to zero
We start with the polynomial:
3x2+4x−4=0
Step 2: Factor the polynomial
To factor the polynomial, we need to express it in the form (ax+b)(cx+d). We can look for two numbers that multiply to 3×−4=−12 (the product of the coefficient of x2 and the constant term) and add up to 4 (the coefficient of x).
The numbers that satisfy these conditions are 6 and −2 because:
6×−2=−12 and 6+(−2)=4
Now, we can rewrite the middle term 4x using these numbers:
3x2+6x−2x−4=0
Step 3: Group the terms
Next, we group the terms:
(3x2+6x)+(−2x−4)=0
Step 4: Factor by grouping
Now we factor out the common factors from each group:
3x(x+2)−2(x+2)=0
We can now factor out (x+2):
(x+2)(3x−2)=0
Step 5: Solve for x
Now, we set each factor equal to zero:
1. x+2=0⇒x=−2
2. 3x−2=0⇒3x=2⇒x=2/3
Thus, the zeroes of the polynomial are:
x=−2 and x=2/3
Step 6: Verify the relations between the zeroes and coefficients
Let α=−2 and β=2/3.
Sum of the roots:
α+β=−2+2/3=−6/3+2/3=−4/3
According to Vieta’s formulas, the sum of the roots is given by −b/a:
−b/a=−4/3
Product of the roots:
αβ=−2×2/3=−4/3
According to Vieta’s formulas, the product of the roots is given by c/a:
c/a=−4/3
Both relations hold true, confirming that our zeroes are correct.
Summary of Zeroes:
The zeroes of the polynomial 3x2+4x−4 are:
x=−2 and x=2/3