halo potrzebuje pomocy na juz!!!

ZAD2 WYZNACZ ILOCZYN
a) (x ² + x - 1)(x² - x + 1)
b) (2x²-3x+2) (x²+4 x-3)
c) (2x² + ½x + 1)(x²-x-¼)
d) (x-1)(x + 2)(x-3)
e)(2x-1)(x+3)(2x-1)
f) (x + ½)( 2x +4)(x+4)​


Odpowiedź :

a)

[tex]( {x}^{2} + x - 1)( {x}^{2} - x + 1) = {x}^{4} - {x}^{3} + {x}^{2} + {x}^{3} - {x}^{2} + x - {x}^{2} + x - 1 = {x}^{4} - {x}^{2} + 2x - 1[/tex]

b)

[tex](2 {x}^{2} - 3x + 2)( {x}^{2} + 4x - 3) = 2 {x}^{4} + 8 {x}^{3} - 6 {x}^{2} - 3 {x}^{3} - 12 {x}^{2} + 9x + 2 {x}^{2} + 8x - 6 = 2 {x}^{4} + 5 {x}^{3} - 16 {x}^{2} + 17x - 6[/tex]

c)

[tex](2 {x}^{2} + \frac{1}{2} x + 1)( {x}^{2} - x - \frac{1}{4}) = 2 {x}^{4} - 2 {x}^{3} - \frac{1}{2} {x}^{2} + \frac{1}{2} {x}^{3} - \frac{1}{2} {x}^{2} - \frac{1}{8}x + {x}^{2} - x - \frac{1}{4} = 2 {x}^{4} - \frac{3}{2} {x}^{3} - \frac{9}{8} x - \frac{1}{4} [/tex]

d)

[tex](x - 1)(x + 2)(x - 3) = ( {x}^{2} + 2x - x - 2)(x - 3) = ( {x}^{2} + x - 2)(x - 3 )= {x}^{3} - 3 {x}^{2} + {x}^{2} - 3x - 2x + 6 = {x}^{3} - 2 {x}^{2} - 5x + 6 [/tex]

e)

[tex](2x - 1)(x + 3)(2x - 1) = (2 {x}^{2} + 6x - x - 3)(2x - 1) = (2 {x}^{2} + 5x - 3)(2x - 1) = 4 {x}^{3} - 2 {x}^{2} + 10 {x}^{2} - 5x - 6x + 3 = 4 {x}^{3} + 8 {x}^{2} - 11x + 3[/tex]

f)

[tex](x + \frac{1}{2})(2x + 4)(x + 4) = (2 {x}^{2} + 4x + x + 2)(x + 4) = (2 {x}^{2} + 5x + 2)(x + 4) = 2 {x}^{3} + 8 {x}^{2} + 5 {x}^{2} + 20x + 2x + 8 = 2 {x}^{3} + 13 {x}^{2} + 22x + 8[/tex]