Whakaoti i te (x-frac{3-sqrt{5}}{2})(x-frac{sqrt{5}+3}{2}

$$( x - \frac { 3 - \sqrt { 5 } } { 2 } ) ( x - \frac { \sqrt { 5 } + 3 } { 2 }$$

Aromātai

$$x^{2}-3x+1$$

Ngā Hipanga Rongoā Poto

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{2}{2}.

$$\left(\frac{2x}{2}-\frac{3-\sqrt{5}}{2}\right)\left(x-\frac{\sqrt{5}+3}{2}\right)$$

Tā te mea he rite te tauraro o \frac{2x}{2} me \frac{3-\sqrt{5}}{2}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{2x-\left(3-\sqrt{5}\right)}{2}\left(x-\frac{\sqrt{5}+3}{2}\right)$$

Mahia ngā whakarea i roto o 2x-\left(3-\sqrt{5}\right).

$$\frac{2x-3+\sqrt{5}}{2}\left(x-\frac{\sqrt{5}+3}{2}\right)$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{2}{2}.

$$\frac{2x-3+\sqrt{5}}{2}\left(\frac{2x}{2}-\frac{\sqrt{5}+3}{2}\right)$$

Tā te mea he rite te tauraro o \frac{2x}{2} me \frac{\sqrt{5}+3}{2}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{2x-3+\sqrt{5}}{2}\times \frac{2x-\left(\sqrt{5}+3\right)}{2}$$

Mahia ngā whakarea i roto o 2x-\left(\sqrt{5}+3\right).

$$\frac{2x-3+\sqrt{5}}{2}\times \frac{2x-\sqrt{5}-3}{2}$$

Me whakarea te \frac{2x-3+\sqrt{5}}{2} ki te \frac{2x-\sqrt{5}-3}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$\frac{\left(2x-3+\sqrt{5}\right)\left(2x-\sqrt{5}-3\right)}{2\times 2}$$

Whakareatia te 2 ki te 2, ka 4.

$$\frac{\left(2x-3+\sqrt{5}\right)\left(2x-\sqrt{5}-3\right)}{4}$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2x-3+\sqrt{5} ki ia tau o 2x-\sqrt{5}-3.

$$\frac{4x^{2}-2x\sqrt{5}-6x-6x+3\sqrt{5}+9+2\sqrt{5}x-\left(\sqrt{5}\right)^{2}-3\sqrt{5}}{4}$$

Pahekotia te -6x me -6x, ka -12x.

$$\frac{4x^{2}-2x\sqrt{5}-12x+3\sqrt{5}+9+2\sqrt{5}x-\left(\sqrt{5}\right)^{2}-3\sqrt{5}}{4}$$

Pahekotia te -2x\sqrt{5} me 2\sqrt{5}x, ka 0.

$$\frac{4x^{2}-12x+3\sqrt{5}+9-\left(\sqrt{5}\right)^{2}-3\sqrt{5}}{4}$$

Ko te pūrua o \sqrt{5} ko 5.

$$\frac{4x^{2}-12x+3\sqrt{5}+9-5-3\sqrt{5}}{4}$$

Tangohia te 5 i te 9, ka 4.

$$\frac{4x^{2}-12x+3\sqrt{5}+4-3\sqrt{5}}{4}$$

Pahekotia te 3\sqrt{5} me -3\sqrt{5}, ka 0.

$$\frac{4x^{2}-12x+4}{4}$$

Whakawehea ia wā o 4x^{2}-12x+4 ki te 4, kia riro ko 1-3x+x^{2}.

$$1-3x+x^{2}$$

Tauwehe

$$\left(x-\frac{3-\sqrt{5}}{2}\right)\left(x-\frac{\sqrt{5}+3}{2}\right)$$