Question

Sa se determine in produs de factori ireductibili peste R polinoamele :
a) f=4x^4-13x^2+9
b)f=4x^3-3x^2-3x+4
c)f=x^3-8
d)f=25x^4-10x^2+1

Answer 1

Răspuns:

a)   x²=y   y>0

4y²-13y+9=0

Δ=13²-4*4*9=169-144=25

√Δ=√25=5

y1=(13-5)2*4=8/8=1

x²=1

x1= -1

x2=1

y2=(13+5)/8=18/8=9/4

x²=9/4

x3=√9/4=-3/2

x4=3/2

f=(x-1)(x+1)*(x+3/2)(x-3/2)

b)4x³-3x²-3x+4=

(4x³+4)-3x(x+1)=

4(x³+1)-3x(x+1)=

4(x+1)(x²-x+1)-3x(x+1)=

4(x+1)(x²-x+1-3x)

4(x+1)(x²-4x+1)

x²-4x+1=0

Δ=(-4)²-4=16-4=12

√Δ=√12=+/-2√3

x1=(4+2√3)/2=2+√3

x2=2-√3

f=4(x+1)(x-2-√3)(x-2+√3)

c)f=x³-8=(x-2)(x²-2x+4)

d)f=25x^4-10x²+1=(5x²-1)²

5x²-1=0

5x²=1

x²=1/5

x1=-√1/5=-√5/5

x2=√5/5

5x²-1=(x+√5/5)(x-√5/5)

f=[(x+√5/5)(x-√5/5)]²

Explicație pas cu pas:

Answer 2

$\it ^{a)\ 4x^4-13x^2+9=4x^4-4x^2-9x^2+9=4x^2(x^2-1)-9(x^2-1)=\\ (x^2-1)(2^2x^2-3^2)=(x-1)(x+1)(2x-3)(2x+3)}\\ \\ \\ ^{b)\ 4x^3-3x^2-3x+4=4(x^3+1)-3x(x+1)=4(x+1)(x^2-x+1)-3x(x+1)=(x+1)(4x^2-4x+4-3x)=(x+1)(4x^2-7x+4)}\\ \\ \\ ^{c)\ x^3-8=x^3-2^3=(x-2)(x^2-2x+4)}\\ \\ \\ ^{d)\ 25x^4-10x^2+1=(5x^2)^2-10x^2+1^2=(5x^2-1)^2}$