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# multiplying polynomials

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can I multiply (2x+5y)(3x^2-4xy+2y^2)?

gadzooks1753  Nov 16, 2017
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### 6+0 Answers

#1
-1

Expand the following:
(2 x + 5 y) (3 x^2 - 4 x y + 2 y^2)

(3 x^2 - 4 x y + 2 y^2) (2 x + 5 y) = 2 x (3 x^2 - 4 x y + 2 y^2) + 5 y (3 x^2 - 4 x y + 2 y^2):
2 x (3 x^2 - 4 x y + 2 y^2) + 5 y (3 x^2 - 4 x y + 2 y^2)

2 x (3 x^2 - 4 x y + 2 y^2) = 2 x×3 x^2 + 2 x (-4 x y) + 2 x×2 y^2:
2×3 x x^2 - 4×2 x x y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

2 x×3 x^2 = 2 x^(1 + 2)×3:
2×3 x^(1 + 2) - 4×2 x x y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

1 + 2 = 3:
2×3 x^3 - 4×2 x x y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

2×3 = 6:
6 x^3 - 4×2 x x y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

2 x (-4) x y = 2 x^2 (-4) y:
6 x^3 + -4×2 x^2 y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

2 (-4) = -8:
6 x^3 + -8 x^2 y + 2×2 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

2×2 = 4:
6 x^3 - 8 x^2 y + 4 x y^2 + 5 y (3 x^2 - 4 x y + 2 y^2)

5 y (3 x^2 - 4 x y + 2 y^2) = 5 y×3 x^2 + 5 y (-4 x y) + 5 y×2 y^2:
6 x^3 - 8 x^2 y + 4 x y^2 + 5×3 y x^2 - 4×5 y x y + 5×2 y y^2

5×3 = 15:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 - 4×5 y x y + 5×2 y y^2

5 y (-4) x y = 5 y^2 (-4) x:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 + -4×5 y^2 x + 5×2 y y^2

5 (-4) = -20:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 + -20 y^2 x + 5×2 y y^2

5 y×2 y^2 = 5 y^(1 + 2)×2:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 - 20 y^2 x + 5×2 y^(1 + 2)

1 + 2 = 3:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 - 20 y^2 x + 5×2 y^3

5×2 = 10:
6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 - 20 y^2 x + 10 y^3

Grouping like terms, 6 x^3 - 8 x^2 y + 4 x y^2 + 15 y x^2 - 20 y^2 x + 10 y^3 = 10 y^3 + (4 x y^2 - 20 x y^2) + (15 x^2 y - 8 x^2 y) + 6 x^3:
10 y^3 + (4 x y^2 - 20 x y^2) + (15 x^2 y - 8 x^2 y) + 6 x^3

4 (x y^2) - 20 (x y^2) = -16 (x y^2):
10 y^3 + -16 x y^2 + (15 x^2 y - 8 x^2 y) + 6 x^3

15 (x^2 y) - 8 (x^2 y) = 7 (x^2 y):
10y^3 - 16xy^2 + 7x^2y + 6x^3

Guest Nov 16, 2017
#2
+80
-1

so i can multiply this

gadzooks1753  Nov 16, 2017
#3
+733
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Yes.

AdamTaurus  Nov 16, 2017
#4
+80
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is the answer 6x^3+7x^2y-16xy^2+10y^3

gadzooks1753  Nov 16, 2017
#5
+80
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nevermind on that answer

gadzooks1753  Nov 16, 2017
#6
+733
0

(2x+5y)(3x^2-4xy+2y^2)

First, multiply the first term in the first parenthesis by every term in the polynomial.

\((2x)(3x^2)+(2x)(-4xy)+(2x)(2y^2)+(5y)(3x^2-4xy+2y^2)\)

Simplify.

\((2x)(3x^2)=6x^3, (2x)(-4xy)=-8x^2y, (2x)(2y^2)=4xy^2\)

\(6x^3-8x^2y+4xy^2+(5y)(3x^2-4xy+2y^2)\)

Now distribute the 5y.

\(6x^3-8x^2y+4xy^2+(5y)(3x^2)+(5y)(-4xy)+(5y)(2y^2)\)

Simplify.

\((5y)(3x^2)=15x^2y, (5y)(-4xy)=-20xy^2, (5y)(2y^2)=10y^3\)

\(6x^3-8x^2y+4xy^2+15x^2y-20xy^2+10y^3\)

Reorganize.

\(6x^3-8x^2y+15x^2y+4xy^2-20xy^2+10y^3\)

Combine like terms.

\(-8x^2y+15x^2y=7x^2y, 4xy^2-20xy^2=-16xy^2\)

\(6x^3+7x^2y-16xy^2+10y^3\)

This is the same as Guest's answer but LaTeX makes it easier to read.

AdamTaurus  Nov 16, 2017

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