Question 19

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials (vi) $4x^2+5\sqrt{2}x-3$.

Solution:

To find the zeroes of the polynomial $$4{\mathrm{x2}}^{}+5\sqrt{2}x-3$$ using the factorization method, we will follow these steps:

Step 1: Set the polynomial equal to zero
We start by setting the polynomial equal to zero:
$$4{\mathrm{x2}}^{}+5\sqrt{2}x-3=0$$

Step 2: Factor the polynomial
We will use the middle-term splitting method to factor the polynomial. We need to express $$5\sqrt{2}x$$ as a sum of two terms whose coefficients multiply to give $$4\times (-3)=-12$$ and add up to $$5\sqrt{2}$$.

We can split $$5\sqrt{2}x$$ into $$6\sqrt{2}x-\sqrt{2}x$$:
$$4{\mathrm{x2}}^{}+6\sqrt{2}x-\sqrt{2}x-3=0$$

Now, we can group the terms:
$$(4{\mathrm{x2}}^{}+6\sqrt{2}x)+(-\sqrt{2}x-3)=0$$

Step 3: Factor by grouping
Now we will factor by grouping:
1. From the first group $$4{\mathrm{x2}}^{}+6\sqrt{2}x$$, we can take out $$2\sqrt{2}x$$:
$$2\sqrt{2}x(2x+3)$$
2. From the second group $$-\sqrt{2}x-3$$, we can factor out $$-1$$:
$$-1(\sqrt{2}x+3)=-1(2x+3)$$

So we have:
$$2\sqrt{2}x(2x+3)-1(2x+3)=0$$

Now we can factor out the common term $$(2x+3)$$:
$$(2x+3)(2\sqrt{2}x-1)=0$$

Step 4: Solve for the zeroes
Now we can set each factor equal to zero:
1. $$2x+3=0$$
$$2x=-3⟹x=-\frac{\mathrm{3/}}{2}$$
2. $$2\sqrt{2}x-1=0$$
$$2\sqrt{2}x=1⟹x=\frac{\mathrm{1/}}{2}=\frac{\sqrt{\mathrm{2/}}}{4}$$

Step 5: Summary of the zeroes
The zeroes of the polynomial $$4{\mathrm{x2}}^{}+5\sqrt{2}x-3$$ are:
$$x_1=-\frac{\mathrm{3/}}{2},{\mathrm{x2}}^{}=\frac{\sqrt{\mathrm{2/}}}{4}$$

Step 6: Verify the relations between the zeroes and coefficients
Let $$x_1$$ and $${\mathrm{x2}}^{}$$ be the zeroes. According to Vieta’s formulas:
– The sum of the zeroes $$x_1$$$$+x2=-\frac{\mathrm{b/}}{a}$$
– The product of the zeroes $$x_1$$$$\cdot x2=\frac{\mathrm{c/}}{a}$$

Here, $$a=4$$, $$b=5\sqrt{2}$$, and $$c=-3$$.

1. Sum of the zeroes:
$$x1+x2=-\frac{\mathrm{3/}}{2}+\frac{\sqrt{\mathrm{2/}}}{4}$$
To find a common denominator:
$$=-\frac{\mathrm{6/}}{4}+\frac{\sqrt{\mathrm{2/}}}{4}=(\frac{-}{6}$$
Now, calculate $$-\frac{\mathrm{b/}}{a}$$:
$$-\frac{5}{\sqrt{\mathrm{2/}}}$$

2. Product of the zeroes:
$$x1x2=-\frac{\mathrm{3/}}{2}\cdot \frac{\sqrt{\mathrm{2/}}}{4}=-\frac{3}{\sqrt{\mathrm{2/}}}$$
Now, calculate $$\frac{\mathrm{c/}}{a}$$:
$$\frac{-}{\mathrm{3/}}$$

Conclusion
Thus, we have verified the relations between the zeroes and the coefficients of the polynomial.