Siema ; ] Proszę o pomoc , prosze o jak najlepsze rozwiązanie i opiszcie dobrze ! legenda : x2=czyli x do kwadratu zarówno jak i jak x3=x do sześcianu itd. : )
Rozłóż podane wielomiany na czynniki
a)W(x)=x3+3x+2x+6
b)w(x)=x3+4x2-4x-16
c)w(x)=x3+3x2-2x-6
d)w(x)=x4+x3+2x+2
e)w(x)=x2+3x-4
f)w(x)-2x2+3x-1
g)w(x)x2+10x+25
h)w(x)=x4-8x2+16
i)w(x)=x3+3x2+2x
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a)W(x)=x³+3x²+2x+6
W(x)=x²(x+3)+2(x+3)
W(x)=(x²+2)(x+3)
b)w(x)=x³+4x²-4x-16
w(x)=x²(x+4)-4(x+4)
w(x)=(x²-4)(x+4)
w(x)=(x+2)(x-2)(x+4)
c)w(x)=x³+3x²-2x-6
w(x)=x²(x+3)-2(x+3)
w(x)=(x²-2)(x+3)
w(x)=(x-√2)(x+√2)(x+3)
d)w(x)=x⁴+x³+2x+2
w(x)=x³(x+1)+2(x+1)
w(x)=(x³+2)(x+1)
e)w(x)=x²+3x-4
w(x)=(x-4)(x-1)
f)w(x)=-2x²+3x-1
w(x)=(x+1)(x+1/2)
g)w(x)=x²+10x+25
w(x)=(x+5)²
h)w(x)=x⁴-8x²+16
w(x)=(x-2)(x³+2x²-4x-8)
w(x)=(x-2)[x²(x+2)-4(x+2)]
w(x)=(x-2)(x²-4)(x+2)
w(x)=(x-2)(x-2)(x+2)(x+2)
i)w(x)=x³+3x²+2x
w(x)=x(x²+3x+2)
w(x)=x(x-2)(x-1)