Question

divide 6x3+√2x2-10x-4√2 by x-√2​

Answer 1

answer:

6x2+7root2x-4

Answer 2

x² - 10 is the quotient and -14√2 is the remainder

Given

• 6x³ + √2x² - 10x - 4√2
• x - √2

To Find

Quotient and remainder by dividing the two numbers

Solution

First, we will write the expressions just as we do in the long-division method

$x - \sqrt{2}\overline{) x^3 - \sqrt{2}x^2 - 10x - 4\sqrt{2}}$

Now we will look for the perfect variable to multiply the first term of the divisor, i.e x to get the first term of the dividend, i.e x³

Since x.x² = x³, our first term for the quotient is x².

Now we will subtract the product from the dividend like this-

$\hspace{1.25cm}x^2\\x - \sqrt{2}\overline{) x^3 - \sqrt{2}x^2 - 10x - 4\sqrt{2}}\\{\hspace{1cm}-x^3+\sqrt{2}x^2}\\{\hspace{1.25cm}\overline{{\hspace{1cm}} 0{\hspace{1cm}}}}$

Now, we will bring -10x - 4√2 down

$\hspace{1.25cm}x^2\\x - \sqrt{2}\overline{) x^3 - \sqrt{2}x^2 - 10x - 4\sqrt{2}}\\{\hspace{1cm}-x^3+\sqrt{2}x^2}\\{\hspace{1.25cm}\overline{{\hspace{1.6cm}}-10x-4\sqrt{2} }}}$

Just like before, we now get -10 in our quotient.

We will subtract -10x + 10√2 to get

$\hspace{1.25cm}x^2-10\\x - \sqrt{2}\overline{) x^3 - \sqrt{2}x^2 - 10x - 4\sqrt{2}}\\{\hspace{1cm}-x^3+\sqrt{2}x^2}\\{\hspace{1.25cm}\overline{{\hspace{1.6cm}}-10x-4\sqrt{2} }}}\\ {\hspace{3.25cm}10x-10\sqrt{2}}\\{\hspace{1.25cm}\overline{{\hspace{2.3cm}}0-14\sqrt{2} }}}$

Since we cannot further multiply anything with x to match up with -14√2,

we get our answer as

x² - 10 as the quotient

-14√2 as the remainder

Therefore, x² - 10 is the quotient and -14√2 is the remainder.

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