Question

Find the product carefully (y^2 + 7) (3y^3 + 2y^2 + y + 7) Read carefully

Answer 1

QUESTION:-

✯ғɪɴᴅ ᴛʜᴇ ᴘʀᴏᴅᴜᴄᴛ of $(ʏ^2 + 7) (3ʏ^3 + 2ʏ^2 + ʏ + 7)$

$\Large\underline\bold{TO \:FIND,}$

$\sf\dashrightarrow the \:product\:of\:given\:equation$

ANSWER✔

$\sf\therefore (ʏ^2 + 7) (3ʏ^3 + 2ʏ^2 + ʏ + 7)$

$\sf\implies (y^2) \times (3ʏ^3 + 2ʏ^2 + ʏ + 7) +(7) \times (3ʏ^3 + 2ʏ^2 + ʏ + 7)$

$\sf\implies \big( 3y^5+2y^4+y^3+7y^2 \big) + \big( 21y^3+14y^2+7y+49 \big)$

$\sf\implies 3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49$

$\sf\therefore the\: product\:obtained\:is ,$

$\sf\therefore 3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49$

$\large{\boxed{\sf{product=3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49}}}$

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✯ADDITIONAL INFORMATION,

FEW IDENTITIES TO LEARN,

$\sf\therefore (a+b)^2=a^2+2ab+b^2$

$\sf\therefore (a+b)(a-b)=a^2-b^2$

$\sf\therefore (a+b)^2+(a-b)^2=2(a^2+b^2)$

$\sf\therefore (a+b)^3=a^3+b^3+3ab(a+b)$

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Answer 2

$\huge\mathcal{\fcolorbox{lime}{black}{\pink{Answer}}}$

$\sf\therefore (ʏ^2 + 7) (3ʏ^3 + 2ʏ^2 + ʏ + 7)$

$\sf\implies (y^2) \times (3ʏ^3 + 2ʏ^2 + ʏ + 7) +(7) \times (3ʏ^3 + 2ʏ^2 + ʏ + 7)$

$\sf\implies \big( 3y^5+2y^4+y^3+7y^2 \big) + \big( 21y^3+14y^2+7y+49 \big)$

$\sf\implies 3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49$

$\sf\therefore the\: product\:obtained\:is ,$

$\sf\therefore 3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49$

$\large{\boxed{\sf{product=3y^5+2y^4+21y^3+y^3+7y^2+14y^2+7y+49}}}$