Question 20

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(vii) $2s^2-(1+2\sqrt{2})s+\sqrt{2}$

Solution:

To find the zeroes of the polynomial $2s^2-(1+2\sqrt{2})s+\sqrt{2}$ using the factorization method, we can follow these steps:

Step 1: Write the polynomial in standard form
The given polynomial is:
$2s^2-(1+2\sqrt{2})s+\sqrt{2}$

Step 2: Set the polynomial equal to zero
We need to solve:
$$2s2-(1+2\surd 2)s+\surd 2=0$$

Step 3: Use the method of splitting the middle term
We need to express the middle term $$-(1+2\sqrt{2})s$$ in a way that allows us to factor the polynomial. We look for two numbers that multiply to $$2\cdot \sqrt{2}$$ (the product of the coefficient of $${\mathrm{s2}}^{}$$ and the constant term) and add up to $$-(1+2\sqrt{2})$$.

The two numbers we are looking for are $$-1$$ and $$-2\sqrt{2}$$ because:
$$-1\cdot -2\sqrt{2}=2\sqrt{2}\text{and}-1+(-2\sqrt{2})=-(1+2\sqrt{2})$$

Step 4: Rewrite the polynomial using these numbers
Now we rewrite the polynomial:
$$2{\mathrm{s2}}^{}-1s-2\sqrt{2}s+\sqrt{2}=0$$

Step 5: Factor by grouping
Group the terms:
$$(2{\mathrm{s2}}^{}-2\sqrt{2}s)+(-1s+\sqrt{2})=0$$
Factor out the common terms:
$$2s(s-\sqrt{2})-1(s-\sqrt{2})=0$$
Now, factor out $$(s-\sqrt{2})$$:
$$(2s-1)(s-\sqrt{2})=0$$

Step 6: Set each factor to zero
Now we solve for $$s$$:
1. $$2s-1=0$$
$$2s=1⟹s=\frac{\mathrm{1/}}{2}$$
2. $$s-\sqrt{2}=0$$
$$s=\sqrt{2}$$

Step 7: State the zeroes of the polynomial
The zeroes of the polynomial are:
$$s=\frac{\mathrm{1/}}{2}\text{and}s=\sqrt{2}$$

Step 8: Verify the relations between the zeroes and coefficients
For a quadratic polynomial $$a{s}^2+bs+c$$, the relationships are:
– Sum of the zeroes ($$\alpha +\beta$$) = $$-\frac{\mathrm{b/}}{a}$$
– Product of the zeroes ($$\alpha \beta$$) = $$\frac{\mathrm{c/}}{a}$$

Here, $$a=2$$, $$b=-(1+2\sqrt{2})$$, and $$c=\sqrt{2}$$.

Calculating the sum:
$$\alpha +\beta =\frac{\mathrm{1/}}{2}+\sqrt{2}$$
Calculating $$-\frac{\mathrm{b/}}{a}$$:
$$-\{\frac{-}{(}=(\frac{1}{+}$$

Calculating the product:
$$\alpha \beta =\frac{\mathrm{1/}}{2}\cdot \sqrt{2}=\frac{\sqrt{\mathrm{2/}}}{2}$$
Calculating $$\frac{\mathrm{c/}}{a}$$:
$$\frac{\sqrt{\mathrm{2/}}}{2}$$

Both relations hold true:
– $$\alpha +\beta =-\frac{\mathrm{b/}}{a}$$
– $$\alpha \beta =\frac{\mathrm{c/}}{a}$$

Conclusion
Thus, the zeroes of the polynomial $$2{\mathrm{ss2}}^{}-(1+2\sqrt{2})s+\sqrt{2}$$ are $$\frac{\mathrm{1/}}{2}$$ and $$\sqrt{2}$$.