Use the FOIL method to simplify the binomials. (y-3) (2y^3-4y+5) show all work
there is no such thing as the foil method, that is a lie made up by math teachers
did the question really say "simplify" and really say "foil"?
@satellite73 I'm pretty sure the "foil" method is real
i am pretty sure it is not
yes this is what the problem said
okay thanks for your help
"foil" is a mnemonic device to remember to do 4 multipications when you have two terms by two terms "first, outer, inner, last'"\[(a+b)(c+d)=ac+ad+bc+bc\]
but here you have not two terms by two terms but two terms by 3 terms, meaning you have to do not four but SIX multiplications, more letters than in "foil"
\[(y-3) (2y^3-4y+5) \] \[=y\times 2y^3+y\times (-4y)+y\times 5-3\times 2y^3-3\times (-4y) -3\times 5\]
you don't really write that of course, just go to \[2y^4-4y^2+5y-6y^3+12y-15\]
then combine like terms
tell your math teacher that foil does not work in this case, too many terms then ask what "simplify" means in this context what it really means is "multiply" there is no such mathematical operation as "simplify"
\[ (\overbrace{\overbrace{\color{orange}a +\underbrace{\color{cornflowerblue}b)(\color{seagreen}c}_{\text{Inside}}}^{\text{First}} +\color{brown}d}^{\text{Outside}}) =%unklerhaukus \overbrace{\color{orange}a\color{seagreen}c}^\text F +\overbrace{\color{orange}a\color{brown}d}^\text O +\underbrace{\color{cornflowerblue}b\color{seagreen}c}_\text I +\underbrace{\color{cornflowerblue}b\color{brown}d}_\text L\\ \qquad\quad \underbrace{\qquad\qquad}_{\text{Last}}\] \[(\color{orange}y\color{cornflowerblue}{-3}) (\color{seagreen}{2y^3-4y}+\color{brown}5) =\overbrace{\color{orange}y\times(\color{seagreen}{2y^3-4y})}^{\text F}+\overbrace{\color{orange}y\times\color{brown}5}^{\text O}+\underbrace{\color{cornflowerblue}{-3}\times(\color{seagreen}{2y^3-4y})}_{\text I}+\underbrace{\color{cornflowerblue}{-3}\times \color{brown}5}_{\text L}%unklerhaukus\]
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