Respuesta :

austint1414

[tex]Since\:Both\:x2\:and\:x^2\:Mean\:The\:Same;[/tex]

[tex]Solution, x^2\left(5x-7\right)\left(3x+2\right)=0\quad :\quad x=0,\:x=\frac{7}{5},\:x=-\frac{2}{3}[/tex]

[tex]Steps:[/tex]

[tex]x^2\left(5x-7\right)\left(3x+2\right)=0\quad :\quad x=0,\:x=\frac{7}{5},\:x=-\frac{2}{3}[/tex]

[tex]\mathrm{Using\:the\:Zero\:Factor\:Principle:}[/tex]

[tex]x=0,\\\mathrm{Solve\:}\:5x-7=0,\\\mathrm{Add\:}7\mathrm{\:to\:both\:sides},\\5x-7+7=0+7,\\\mathrm{Simplify},\\5x=7,\\\mathrm{Divide\:both\:sides\:by\:}5,\\\frac{5x}{5}=\frac{7}{5},\\\mathrm{Simplify},\\x=\frac{7}{5};\\\mathrm{Solve\:}\:3x+2=0,\\\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides},\\3x+2-2=0-2,\\\mathrm{Simplify},\\3x=-2,\\\mathrm{Divide\:both\:sides\:by\:}3,\\\frac{3x}{3}=\frac{-2}{3},\\\mathrm{Simplify},\\x=-\frac{2}{3}[/tex]

[tex]\mathrm{The\:final\:solutions\:to\:the\:equation\:are:},\\x=0,\:x=\frac{7}{5},\:x=-\frac{2}{3}[/tex]

[tex]\mathrm{Hope\:This\:Helps!!!}[/tex]

[tex]\mathrm{-Austint1414}[/tex]

According to the factor theorem, the roots are:

[tex]x = 0[/tex]

[tex]x = \frac{7}{5}[/tex]

[tex]x = -\frac{2}{3}[/tex]

The factor theorem states that a polynomial with roots [tex]x_1, x_2, ..., x_n[/tex] can be written as:

[tex](x - x_1)(x - x_2)...(x - x_n) = 0[/tex]

In this problem:

[tex]x^2(5x - 7)(3x + 2) = 0[/tex]

Thus, the roots are:

[tex]x^2 = 0 \rightarrow x = 0[/tex]

[tex]5x - 7 = 0 \rightarrow x = \frac{7}{5}[/tex]

[tex]3x + 2 = 0 \rightarrow x = -\frac{2}{3}[/tex]

A similar problem is given at https://brainly.com/question/24380382

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