Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 346

$$\boxed{{\displaystyle \mathbf{346}.}}$$

$$1)P(x,y)=15x^4-8x^{3}y+$$

$$+31x^2{y}^2-16x{y}^3+2y^4$$

$$y=tx:$$

$$P(x,y)=15x^4-8t{x}^4+$$

$$+31t^2{x}^4-16t^3{x}^4+2t^4{x}^4=$$

$$=x^4(2t^4-16t^3+31t^2-8t+15).$$

$$Q(x,y)=$$

$$=x^4(t-3)(t-5)(2t^2+1).$$

$$Делаемобратнуюзамену:$$

$$\text{ }t=\frac{y}{x}.$$

$$P(x,y)=$$

$$=x^4(\frac{y}{x}-3)(\frac{y}{x}-5)$$

$$(2\cdot \frac{y^2}{x^2}+1)=$$

$$=(y-3x)(y-5x)(2y^2+x^2).$$

$$2)P(x,y)=12x^5-32x^{4}y+$$

$$+9x^3{y}^2+16x^2{y}^3-$$

$$-3x{y}^4-2y^5.$$

$$y=tx:$$

$$P(x,y)=12x^5-32x^{4}tx+$$

$$+9x^3{\left(\text{tx}\right)}^2+16x^2{\left(\text{tx}\right)}^3-$$

$$-3x{(tx)}^4-2{(tx)}^5.$$

$$=12x^5-32x{t}^5+9t^2{x}^5+$$

$$+16t^3{x}^5-3t^4{x}^5-2t^5{x}^5=$$

$$=x^5(-2t^5-3t^4+16t^3+9t^2-32t+12).$$

$$Q(x,y)=(t-1)(t+3)$$

$$(t-2)(t+2)(-2t+1).$$

$$Делаемобратнуюзамену$$

$$t=\frac{y}{x}:$$

$$P(x,y)=x^5(\frac{y}{x}-1)(\frac{y}{x}+3)$$

$$(\frac{y}{x}-2)(\frac{y}{x}+2)(-\frac{2y}{x}+1)=$$

$$=(y-x)(y+3x)(y-2x)$$

$$(y+2x)(-2y+x).$$