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frutty [35]
3 years ago
9

How would you describe your ability to solve equations with two or more steps

Mathematics
1 answer:
dexar [7]3 years ago
3 0
I would describe it as talented and accurate. I have not gotten anything lower than a b+ in any of my classes, including math.
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Consider the following division of polynomials.
Bond [772]

x^4=x^2\cdot x^2. Multiplying the denominator by x^2 gives

x^2(x^2+2x+8)=x^4+2x^3+8x^2

Subtracting this from the numerator gives a remainder of

(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8

-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

=x^2-x+1

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

3 0
2 years ago
Read 2 more answers
Simplify each expression -5r (3r^2)​
Evgen [1.6K]

Answer:

The answer you're looking for is

-5r (3r^2) = -45

Step-by-step explanation:

7 0
3 years ago
72 = 8 x 9
vivado [14]
8988989889 IDK hate doing the it but luv getting the answers
8 0
2 years ago
Read 2 more answers
8x+10y=<br> 8x+10y=<br> \,\,70<br> 70
notsponge [240]

Answer:

54

Step-by-step explanation:

4 0
3 years ago
57/4 as a mixed number still working on my test please help (if yuo vould go look at my last question adn help that would be gre
Harrizon [31]

Answer:

14 1/4

Step-by-step explanation:

Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.

5 0
3 years ago
Read 2 more answers
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