andriod09
Factor trinominal times a binomial
equation in the comments



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andriod09
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\[\left( 2x+4y+7 \right)\left( 5x+8 \right)\]

andriod09
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@ganeshie8

ganeshie8
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is it 'factor' or 'multiply' ?

andriod09
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you have to factor it. but its presented as a multiplication question.

ganeshie8
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hmm interesting.. its already in factored form i guess we cant factor it any further

andriod09
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i think it might be multiplying it though. the question is:
7.How about multiplying a trinomial times a binomial?

ganeshie8
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yea this makes sense :)

ganeshie8
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just have to multiply..

ganeshie8
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you familiar wid multuplying these right ?

andriod09
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somewhat

ganeshie8
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\((2x+4y+7)(5x+8)\)
\((\color{green}{2x}+4y+7)(\color{green}{5x+8})\)

ganeshie8
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first distribute 2x

ganeshie8
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then 4y
and lastly 7

andriod09
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in between them, would i use ()+()+() or just ()()()?

ganeshie8
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thats right ! in between we add. () + () + ()

ganeshie8
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\((\color{green}{2x}+4y+7)(\color{green}{5x+8}) \)
\(10x^2 + 16x + 20xy + 32y + 35x + 56 \)

ganeshie8
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see if above makes sense

andriod09
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\[\left( 10x^{2}+16x \right)+\left( 20xy+24y \right)+\left( 35x+56? \right)\]

ganeshie8
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thats almost right !! except for \(24y\)

andriod09
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i mistyped it. lool. its (20xy+32)

ganeshie8
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right ! now remove ()

ganeshie8
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and see if any like terms are there

andriod09
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16x and 35x and thats it.

ganeshie8
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yup :)

ganeshie8
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add them and write it in order

andriod09
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\[(10x^{2}+51x+20xy+24y+56)\]

ganeshie8
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\(\left( 10x^{2}+16x \right)+\left( 20xy+32y \right)+\left( 35x+56? \right) \)
\(10x^{2}+16x +20xy+32y +35x+56 \)
\(10x^{2}+20xy +51x+32y + 56 \)

ganeshie8
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again 24y. ...

ganeshie8
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if we ignore that 24y mistype, your final polynomial looks good except for one thing

andriod09
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\[(10x^{2}+51x+20xy+32y+56)\]

ganeshie8
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20xy order is more

ganeshie8
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so you have to move it to left,

ganeshie8
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and 51x to right.. swap 20xy and 51x