Question 21

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(viii) $v^2+4\sqrt{3}v-15$

Solution:

To find the zeroes of the polynomial $${\mathrm{v2}}^{}+4\sqrt{3}v-15$$ using the factorization method, we can follow these steps:

Step 1: Set the polynomial equal to zero
We start by setting the polynomial equal to zero:
$${\mathrm{v2}}^{}+4\sqrt{3}v-15=0$$

Step 2: Factor the polynomial
To factor the polynomial, we need to express $$4\sqrt{3}v$$ as a sum of two terms whose product is equal to $$-15$$ (the constant term) and whose sum is equal to $$4\sqrt{3}$$ (the coefficient of $$v$$).

We can rewrite $$4\sqrt{3}v$$ as $$5\sqrt{3}v-\sqrt{3}v$$:
$${\mathrm{v2}}^{}+5\sqrt{3}v-\sqrt{3}v-15=0$$

Step 3: Group the terms
Now, we group the terms:
$$({\mathrm{v2}}^{}+5\sqrt{3}v)+(-\sqrt{3}v-15)=0$$

Step 4: Factor by grouping
Next, we factor out common terms from each group:
$$v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})=0$$
Now we can factor out $$(v+5\sqrt{3})$$:
$$(v-\sqrt{3})(v+5\sqrt{3})=0$$

Step 5: Set each factor to zero
Now, we set each factor equal to zero to find the zeroes:
1. $$v-\sqrt{3}=0$$ → $$v=\sqrt{3}$$
2. $$v+5\sqrt{3}=0$$ → $$v=-5\sqrt{3}$$

Step 6: List the zeroes
The zeroes of the polynomial $${\mathrm{v2}}^{}+4\sqrt{3}v-15$$ are:
$$v=\sqrt{3}\text{and}v=-5\sqrt{3}$$

Step 7: Verify the relations between the zeroes and the coefficients
Let $$\alpha =\sqrt{3}$$ and $$\beta =-5\sqrt{3}$$.

1. Sum of the zeroes:
$$\alpha +\beta =\sqrt{3}-5\sqrt{3}=-4\sqrt{3}$$
According to the relation, $$-\frac{\mathrm{b/}}{a}=-\frac{4}{\sqrt{\mathrm{3/}}}=-4\sqrt{3}$$.

2. Product of the zeroes:
$$\alpha \cdot \beta =\sqrt{3}\cdot (-5\sqrt{3})=-5\cdot 3=-15$$
According to the relation, $$\frac{c}{a}=\frac{-}{\mathrm{15/}}=-15$$.

Both relations are verified, confirming that the zeroes are correct.

Final Answer:
The zeroes of the polynomial $${\mathrm{v2}}^{}+4\sqrt{3}v-15$$ are $$\sqrt{3}$$ and $$-5\sqrt{3}$$.