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Mathematics 8 Online
OpenStudy (anonymous):

Need help distributing f(x)=(x-3)(x-2) and also h(x)= (x-2i)(x+2i)

OpenStudy (anonymous):

Think FOIL: First, Outer, Inner, Last. For f(x): First two terms: (x)*(x)=x^2 Outer two terms: (x)*(-2)= -2x Inner two terms: (-3)*(x)= -3x Last two terms: (-3)*(-2)= 6 Add them all up: x^2-2x-3x+6= x^2-5x+6 Same thing with h(x):=(x-2i)(x+2i) F: (x)*(x)=x^2 O: (x)*(2i)=2ix I: (-2i)*(x)=-2ix L: (-2i)*(2i)=(-4)(i^2) i is the square root of -1, so (-4)*(i^2)=4 Therefore h(x) := x^2+4

OpenStudy (anonymous):

thank you so much :), I'm knew to this, needed help really bad. Do you mind if I ask another question?

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