Question 26

If $\sqrt{2}$ is a zero of $p\left(x\right)=6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2},$ find the remaining zeros

Solution:

To find the remaining zeros of the polynomial $$p\left(x\right)=6x3+\surd 2x2-10x-4\surd \sqrt{2}$$ given that $$\sqrt{2}$$ is one of its zeros, we will use polynomial long division to divide $$p\left(x\right)$$ by $$x-\sqrt{2}$$.

Step 1: Set up the long division
We will divide $$p\left(x\right)$$ by $$x-\sqrt{2}$$.

Step 2: Divide the leading term
The leading term of $$p\left(x\right)$$ is $$6{\mathrm{x3}}^{}$$. We divide this by the leading term of $$x-\sqrt{2}$$, which is $$x$$:
$$\frac{6}{{\mathrm{x3/}}^{}}=6{\mathrm{x2}}^{}$$

Step 3: Multiply and subtract
Now, multiply $$6x^2$$ by $$x-\sqrt{2}$$:
$$6{\mathrm{x2}}^{}(x-\sqrt{2})=6{\mathrm{x3}}^{}-6\sqrt{2}{\mathrm{x2}}^{}$$
Subtract this from $$p\left(x\right)$$:
$$(6{\mathrm{x2}}^{}+\sqrt{2}{\mathrm{x2}}^{}-10x-4\sqrt{2})-(6{\mathrm{x3}}^{}-6\sqrt{2}{\mathrm{x2}}^{})=(1+6)\sqrt{2}{\mathrm{x2}}^{}-10x-4\sqrt{2}=7\sqrt{2}{\mathrm{x2}}^{}-10x-4\sqrt{2}$$

Step 4: Repeat the process
Now, we take the new polynomial $$7\sqrt{2}{\mathrm{x2}}^{}-10x-4\sqrt{2}$$ and divide the leading term $$7\sqrt{2}{\mathrm{x2}}^{}$$ by $$x$$:
$$\frac{7}{\sqrt{2}}=7\sqrt{2}x$$
Multiply $$7\sqrt{2}x$$ by $$x-\sqrt{2}$$:
$$7\sqrt{2}x(x-\sqrt{2})=7\sqrt{2}{\mathrm{x2}}^{}-14$$
Subtract:
$$(7\sqrt{2}{\mathrm{x2}}^{}-10x-4\sqrt{2})-(7\sqrt{2}{\mathrm{x2}}^{}-14)=-10x+14-4\sqrt{2}$$

Step 5: Continue dividing
Now, divide $$-10x+14-4\sqrt{2}$$ by $$x$$:
$$\frac{-}{10}=-10$$
Multiply $$-10$$ by $$x-\sqrt{2}$$:
$$-10(x-\sqrt{2})=-10x+10\sqrt{2}$$
Subtract:
$$(-10x+14-4\sqrt{2})-(-10x+10\sqrt{2})=14-4\sqrt{2}-10\sqrt{2}=14-14\sqrt{2}$$

Step 6: Result of the division
After performing the long division, we find:
$$p\left(x\right)=(x-\sqrt{2})(6{\mathrm{x2}}^{}+7\sqrt{2}x-10)$$

Step 7: Find the remaining zeros
Now we need to find the zeros of the quadratic $$6{\mathrm{x2}}^{}+7\sqrt{2}x-10=0$$ using the quadratic formula:
$$x=[\frac{-}{b}$$
Here, $$a=6$$, $$b=7\sqrt{2}$$, and $$c=-10$$.

Step 8: Calculate the discriminant
Calculate $$b^2-4ac$$:
$$(7\sqrt{2}{)2}^{}-4\cdot 6\cdot (-10)=98+240=338$$

Step 9: Substitute into the quadratic formula
Now substitute into the quadratic formula:
$$x=(\frac{-}{7}$$

Step 10: Final answers
Thus, the remaining zeros of the polynomial $$p\left(x\right)$$ are:
$$x_1=(\frac{-}{7},x_2=(\frac{-}{7}$$