Question

tennetiraj86 please can you help me everyday because I am also teacher from school​

Answer 1

Step-by-step explanation:

Solutions :-

vii)

Given that

(5x-y)(3x²+4xy+y²)

=> 5x(3x²+4xy+y²)-y(3x²+4xy+y²)

=> (5x×3x²)+(5x×4xy)+(5x×y²)-(y×3x²)-(y×4xy)-(y×y²)

=> 15x³+20x²y+5xy²-3x²y-4xy²-y³

=> 15x³+(20x²y-3x²y)+(5xy²-4xy²)-y³

=> 15x³+17x²y+xy²-y³

(5x-y)(3x²+4xy+y²) = 15x³+17x²y+xy²-y³

viii)

Given polynomial is 6p⁴-5p³+8p²-4p+7

Given divisor = (p+1)

6p⁴-5p³+8p²-4p+7 ÷ (p+1)

p+1) 6p⁴-5p³+8p²-4p+7 (6p³-p²+9p-13

6p⁴+6p³

(-) (-)

_______________

-p³+8p²

-p³-p²

(+) (+)

________________

9p²-4p

9p²+9p

(-) (-)

_________________

-13p+7

-13p-13

(+) (+)

__________________

20

___________________

Quotient = 6p³-p²+9p-13

Remainder = 20

ix)

Given equation is [(y+1)-(2y+3)]/(3-2y) = 1/4

=> (y+1-2y-3)/(3-2y) = 1/4

=> (-y-2)/(3-2y) = 1/4

On applying cross multiplication then

=> 4(-y-2) = 1(3-2y)

=> -4y-8 = 3-2y

=> -4y+2y = 3+8

=> -2y = 11

=> y = 11/-2

Therefore y = -11/2

The value of y for the given problem is -11/2