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Question Number 86586 by Power last updated on 29/Mar/20

Commented by Power last updated on 29/Mar/20

$$({\mathrm{x}}_1^5-20)({3\mathrm{x}}_2^4-{2\mathrm{x}}_2-35)=?$$

Commented by MJS last updated on 29/Mar/20

$$-1063$$$$\mathrm{just}\mathrm{solve}\mathrm{and}\mathrm{calculate}$$

Commented by jagoll last updated on 29/Mar/20

$$\mathrm{wow}...\mathrm{super}\mathrm{easy}$$

Commented by Power last updated on 29/Mar/20

$$\mathrm{prove}\mathrm{that}\mathrm{sir}$$

Answered by lémùst last updated on 29/Mar/20

$$x^2=x+3\Rightarrow x^4=x^2+6x+9$$$$\Rightarrow x^4=x+3+6x+9$$$$\Rightarrow x^4=7x+12$$$$\Rightarrow x^5=7x^2+12x=7(x+3)+12x=19x+21$$$$$$$$(x_1^5-20)(3x_2^4-2x_2-35)$$$$=(19x_1+21-20)(21x_2+36-2x_2-35)$$$$=(19x_1+1)(19x_2+1)$$$$={19}^2{x}_1{x}_2+19(x_1+x_2)+1$$$$={19}^2\times (-3)+19\times 1+1$$$$=-1063$$

Commented by Power last updated on 29/Mar/20

$$\mathrm{thanks}$$

Commented by lémùst last updated on 29/Mar/20

$${you}^{\prime }rewelcome$$

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