Question


Answer

4x^3-31x+15

=(4x^2+ax+b)(x+c)

=4x^3+4cx^2+ax^2+acx+bx+bc

=4x^3+(4c+a)x^2+(ac+b)x+bc

\therefore \begin{cases} 4c+a=0 \\ ac+b=-31 \\ bc=15 \end{cases}

\begin{cases} -4c^2+b=-31 \\ bc=15 \end{cases}

-4c^2+\dfrac{15}{c} =-31

4c^3-31c-15=0

\therefore c=3

b=5

a=-12

4x^3-31x+15

=(4x^2+ax+b)(x+c)

=(4x^2-12x+5)(x+3)

=(2x-1)(2x-5)(x+3)

Steven Zheng posted 1 year ago

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