Writing in standard form

Question

I have the following:

$$-11(0.1x-0.2)(0.3x-0.4)$$

I know that the answer is

$$−0.33x^2+1.1x−0.88 $$

But what steps do i need to take to write it in standard form? I don't fully grasp the steps, so if anybody would like explain them to me I'd feel blessed.

Note that $-11(0.1x-0.2)(0.3x-0.4)=-0.33x^2+1.1x-0.88$. So either the original problem, or the answer you wrote, (or both) have a typo. — user729424, Mar 01 '20 at 22:05

score 0 · Accepted Answer · answered Mar 01 '20 at 22:15

$-11(0.1x-0.2)(0.3x-0.4)$

Use the distributive property.

Where do you want to distribute first?

How about breaking up $(0.1x) + (-0.2)$

The other factors multiply by each of these terms.

$-11(0.1x)(0.3x-0.4)+(-11)(-0.2)(0.3x-0.4)$

Then we can multiply the $(-11)$ and $(-0.1x)$ factors and the $(-11)$ and $(-0.2)$ factors.

$(-1.1x)(0.3x-0.4)+(0.22)(0.3x-0.4)$

Keep distributing.

$(-1.1x)(0.3x)+ (-1.1x)(-0.4)+(0.22)(0.3x) + (0.22)(-0.4)$

Multiply

$-3.3x^2 + 0.44x +0.66x - 0.88$

Combine the $x$ terms.

$-3.3x^2 + 1.1x - 0.88\\ $

Writing in standard form

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