multiply (u+1) (u-5)
(a+b)(c+d) a(c+d)+b(c+d) use distributive property
\((\overbrace{\overbrace{\color{orange}a +\underbrace{\color{cornflowerblue}b)(\color{seagreen}c}_{\text{Inside}}}^{\text{First}} +\color{brown}d}^{\text{Outside}}) =\overbrace{\color{orange}a\color{seagreen}c}^\text F +\overbrace{\color{orange}a\color{brown}d}^\text O +\underbrace{\color{cornflowerblue}b\color{seagreen}c}_\text I +\underbrace{\color{cornflowerblue}b\color{brown}d}_\text L\\ \qquad\quad \underbrace{\qquad\qquad}_{\text{Last}}\)
I don't understand , so its u(u-5) + 1(u-5) ? @myininaya
yes @Kb1997
I got u^2-5u+1?? @myininaya
so u(u-5)=u^2-5u right? and 1(u-5)=u-5 so add those two together that is add u^2-5u and u-5 u^2-5u+u-5=?
4u?
u^2-5u+u-5 -5u and u are the only like terms you got there -5u+u is not 4u but is it ?
-5+1=-4 so -5u+1u=-4u right? so you have u^2-5u+u-5 = u^2-4u-5
factor y^2-10y+16 @myininaya
what are two factors of 16 that have product 16 and sum -10?
8 , 2
@myininaya ???
Join our real-time social learning platform and learn together with your friends!