how to solve it.....plzz tell me

Subject:
Math
Author:
braydon5wle
Created:
1 year ago

Answers 2

Answer:

Step-by-step explanation:

3.(2y+1)×(y2−2y3+3y)

=2y(y2−2y3+3y)+1(y2−2y3+3y)

=2y^3−4y^4+6y^2+y^2−2y^3+3y

=−4y^4+2y^3−2y^3+6y^2+y^2+3y

=−4y^4+7y^2+3y

Always use distributive property for solving these kinds of questions.

Author:

loosetoothkfik
Rate an answer:
10

Answer:

(i) 2x³-4x²-2x(ii) x⁸ -x³ + 2x⁷ -2x² +2x⁵-2

(iii) 2y³+y²-2y⁴-2y³+6y²+3y

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*note: when multiplying two same variables,always add the powers.......

for eg: x³ × x⁴ = x⁷*

Step-by-step explanation:

(i) 2x; x²-2x-1

Step : multiply 2x with each term in the other polynomial....

2x × x²-2x -1 = (2x × x²) + (2x × -2x) + (2x × -1)

= 2x3 - 4x² - 2x

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(ii) x⁵-1; x³+2x²+2

Step 1 : multiply x⁵ with each term in the other polynomial....

x⁵ × x³+2x²+2 = (x⁵ × x³) + (x⁵ × 2x²) + (x⁵ × 2)

=. x⁸ + 2x⁷ + 2x⁵ ----> (a)

Step 2: multiply -1 with each term in the other polynomial.......

-1 × x³+2x²+2 = (-1 × x³) + (-1 × 2x²) + (-1 × 2)

= -x³-2x²-2 ------->(b)

Step 3: now add the two products....

(x⁸ + 2x⁷ + 2x⁵) + (-x³-2x²-2)

=x⁸+2x⁷+2x⁵-x³-2x²-2

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Let's follow the same steps to solve part (iii)...

(iii) 2y +1 ; y² - 2y³ + 3y

2y × y² - 2y³ + 3y = (2y × y²) + (2y × 2y³) + (2y × 3y)

= 2y³ + 4y⁴ + 6y²

1 × y² - 2y³ + 3y = (1 × y²) + (1 × 2y³) + ( × 3y)

= y² - 2y³ + 3y

(2y³ + 4y⁴ + 6y²) + (y² - 2y³ + 3y)

= 2y³ - 2y³ + 4y⁴ + 6y² -y² +3y

= 4y⁴ + 6y² -y² +3y

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Thanks for the question

Author:

deven
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