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Writing a polynomial in standard form? PLEASE help?

Writing a polynomial in standard form? PLEASE help? Topic: How to write a polynomial in standard form
June 19, 2019 / By Shelumiel
Question: x(x+2)^2 (Note that ^2 is the "squared") I'm confused on how to do it this way, if you can help i would really appreciate it! Thank You!
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Best Answers: Writing a polynomial in standard form? PLEASE help?

Norman Norman | 1 day ago
Just remember PEDMAS - Parenthesis, Exponents, Division, Mulitplication, Addition, Subtraction (or BEDMAS depending if you know the difference between parenthesis and brackets). Standard form is like x^2 + x + 2 = 0. So first distribute the square (Exponent first, since the brackets can't be evaluated) Now you have: x(x + 2)(x + 2) Parenthesis: Multiply out with FOIL x * (x^2 + 2x + 2x + 4) Now on to multiplication: The x has to be distributed to all terms inside the brackets, so. x^2 * x = x^3 x * 4x = 4x^2 x * 4 = 4x So your equation in standard form is: x^3 + 4x^2 + 4x
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Norman Originally Answered: Writing a polynomial in standard form? PLEASE help?
Just remember PEDMAS - Parenthesis, Exponents, Division, Mulitplication, Addition, Subtraction (or BEDMAS depending if you know the difference between parenthesis and brackets). Standard form is like x^2 + x + 2 = 0. So first distribute the square (Exponent first, since the brackets can't be evaluated) Now you have: x(x + 2)(x + 2) Parenthesis: Multiply out with FOIL x * (x^2 + 2x + 2x + 4) Now on to multiplication: The x has to be distributed to all terms inside the brackets, so. x^2 * x = x^3 x * 4x = 4x^2 x * 4 = 4x So your equation in standard form is: x^3 + 4x^2 + 4x
Norman Originally Answered: Writing a polynomial in standard form? PLEASE help?
It's simple really. Since the x+2 inside of the perenthisis is squared you need to re-write it like so: X(X+2)(X+2) Now, distribute the X that is outside of the perenthisis, leaving you with: x^2+2x(x+2) Now distribute the 2x to the terms inside of the perenthisis. X^2+2x^2+4x Combine like terms and your answer is: 3x^2+4x Hope this helps =D

Kyran Kyran
It's simple really. Since the x+2 inside of the perenthisis is squared you need to re-write it like so: X(X+2)(X+2) Now, distribute the X that is outside of the perenthisis, leaving you with: x^2+2x(x+2) Now distribute the 2x to the terms inside of the perenthisis. X^2+2x^2+4x Combine like terms and your answer is: 3x^2+4x Hope this helps =D
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Kyran Originally Answered: Slope and Slope Intercepts into Standard Form?
You mean: slope = 3/4 ... y = mx + b or y = (3/4) x + b ← plug in slope or 3 = (3/4)(-2) + b ← plug in point or 3 + 3/2 = b or b = 9/2 ... y = (3/4)x + (9/2) or 4y = 3x + 18 or 3x - 4y = - 18 ← ans d.
Kyran Originally Answered: Slope and Slope Intercepts into Standard Form?
you must have omitted a / so I am assuming the slope is 3/4 using the point slope form of a line with slope m and through point (h,k) y-k = m(x-h) for m = 3/4 and pt(-2,3) y-3 = 3/4(x +2) simplify 4y-12 = 3x +6 3x - 4y = -18 d.

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