x^3left(x+2right)^2left(x-1right)left(x-2right) eching

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

Baholash

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

Yechim qadamlari

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

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

x^{3} ga x^{2}+4x+4 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(x^{5}+4x^{4}+4x^{3}\right)\left(x-1\right)\left(x-2\right)$$

x^{5}+4x^{4}+4x^{3} ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

x^{6}+3x^{5}-4x^{3} ga x-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{7}+x^{6}-6x^{5}-4x^{4}+8x^{3}$$

Kengaytirish

$$x^{7}+x^{6}-6x^{5}-4x^{4}+8x^{3}$$

Yechim qadamlari

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

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

x^{3} ga x^{2}+4x+4 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(x^{5}+4x^{4}+4x^{3}\right)\left(x-1\right)\left(x-2\right)$$

x^{5}+4x^{4}+4x^{3} ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

x^{6}+3x^{5}-4x^{3} ga x-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{7}+x^{6}-6x^{5}-4x^{4}+8x^{3}$$