(x-1)(x+2)^2(x-i)^2 eching

$$( x - 1 ) ( x + 2 ) ^ { 2 } ( x - i ) ^ { 2 }$$

Baholash

$$\left(x-1\right)\left(\left(x+2\right)\left(x-i\right)\right)^{2}$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x-1\right)\left(x^{2}+4x+4\right)\left(x-i\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x-i\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x-1\right)\left(x^{2}+4x+4\right)\left(x^{2}-2ix-1\right)$$

x-1 ga x^{2}+4x+4 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(x^{3}+3x^{2}-4\right)\left(x^{2}-2ix-1\right)$$

x^{3}+3x^{2}-4 ga x^{2}-2ix-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{5}+\left(3-2i\right)x^{4}+\left(-1-6i\right)x^{3}-7x^{2}+8ix+4$$

Kengaytirish

$$x^{5}+\left(3-2i\right)x^{4}+\left(-1-6i\right)x^{3}-7x^{2}+8ix+4$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x-1\right)\left(x^{2}+4x+4\right)\left(x-i\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x-i\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x-1\right)\left(x^{2}+4x+4\right)\left(x^{2}-2ix-1\right)$$

x-1 ga x^{2}+4x+4 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(x^{3}+3x^{2}-4\right)\left(x^{2}-2ix-1\right)$$

x^{3}+3x^{2}-4 ga x^{2}-2ix-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{5}+\left(3-2i\right)x^{4}+\left(-1-6i\right)x^{3}-7x^{2}+8ix+4$$