Mathematics, 10.11.2020 03:00 katieleeisaacs8368

Explain the steps that you take to complete an integer long division problem. Then explain the steps in completing a polynomial long division problem. How are they the same? How
are they different? Be sure to fully answer the question clearly and concisely.

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Answer from: reyamukhtar

Step-by-step explanation:

Divide the tens column dividend by the divisor.

Multiply the divisor by the quotient in the tens place column.

Subtract the product from the divisor.

Bring down the dividend in the ones column and repeat.

Purplemath

If you're dividing a polynomial by something more complicated than just a simple monomial (that is, by something more complicated than a one-term polynomial), then you'll need to use a different method for the simplification. That method is called "long polynomial division", and it works just like the long (numerical) division you did back in elementary school, except that now you're dividing with variables.

Divide x2 – 9x – 10 by x + 1

Think back to when you were doing long division with plain old numbers. You would be given one number (called the divisor) that you had to divide into another number (called the dividend). You set up the long-division symbol, inserted the two numbers where they belonged, and then started making guesses as to what should go on top of the symbol.

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And you didn't guess the whole answer right away; instead, you started working on the "front" part (that is, the larger place-value part) of the number you were dividing. For instance, if you were dividing 1137 by 82, you'd look at the "8" and the "10", and guess that probably a "1" should go on top, above the "11", because 8 fits once into 11.

Long division for polynomials works in much the same way:

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

Answer from: Quest

answer is 12345678900987654321234567890987654321

step-by-step explanation:

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Explain the steps that you take to complete an integer long division problem. Then explain the step...