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If the solutions of p(x)=0 are -14 and 11, which function could be p?

A. p(x)=[tex]x^{2}[/tex]-3x-154

B. p(x)=[tex]x^{2}[/tex]-14x+11

C. p(x)=[tex]x^{2}[/tex]+14x+11

D. p(x)=[tex]x^{2}[/tex]+3x-154

Respuesta :

hnori22003

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Solution :

[tex]x = - 14[/tex]

[tex]x = 11[/tex]

Thus ;

[tex]x+ 14 = 0[/tex]

[tex]x - 11 = 0[/tex]

Multiply above equations :

[tex](x + 14)(x - 11) = 0[/tex]

[tex] {x}^{2} + ( 14 - 11)x +( \: ( 14) \times ( - 11) \: ) = 0 \\ [/tex]

[tex] {x}^{2} + 3x - 154 = 0[/tex]

So ;

[tex]p(x) = {x}^{2} + 3x - 154[/tex]

Thus the correct answer is (( D )) .

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