F(x)=(x)^3(x+4)^2(x-1)(x-2) eching

$$F ( x ) = ( x ) ^ { 3 } ( x + 4 ) ^ { 2 } ( x - 1 ) ( x - 2 )$$

Baholash

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

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+4\right)^{2} kengaytirilishi uchun ishlating.

$$x^{3}\left(x^{2}+8x+16\right)\left(x-1\right)\left(x-2\right)$$

x^{3} ga x^{2}+8x+16 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(x^{5}+8x^{4}+16x^{3}\right)\left(x-1\right)\left(x-2\right)$$

x^{5}+8x^{4}+16x^{3} ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(x^{6}+7x^{5}+8x^{4}-16x^{3}\right)\left(x-2\right)$$

x^{6}+7x^{5}+8x^{4}-16x^{3} ga x-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{7}+5x^{6}-6x^{5}-32x^{4}+32x^{3}$$

Kengaytirish

$$x^{7}+5x^{6}-6x^{5}-32x^{4}+32x^{3}$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+4\right)^{2} kengaytirilishi uchun ishlating.

$$x^{3}\left(x^{2}+8x+16\right)\left(x-1\right)\left(x-2\right)$$

x^{3} ga x^{2}+8x+16 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(x^{5}+8x^{4}+16x^{3}\right)\left(x-1\right)\left(x-2\right)$$

x^{5}+8x^{4}+16x^{3} ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(x^{6}+7x^{5}+8x^{4}-16x^{3}\right)\left(x-2\right)$$

x^{6}+7x^{5}+8x^{4}-16x^{3} ga x-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{7}+5x^{6}-6x^{5}-32x^{4}+32x^{3}$$