Find the values of k that make \( 2x^{2}+kx+7 \) factorable.

The values of k that make >      2x² + kx + 7     factorable
the values of k are      9 & 15
thus
(2x² + 9x + 7)  =      (2x + 7)(x+ 1)              k -9
(2x²  + 15x + 7)  =    (2x +1)(x+7)                K  =15

\( 1)\\(2x+1)(x+7)=2x^2+14x+x+7=2x^2+15x+7\ \ \Rightarrow\ \ k=15\\\\2)\\(2x-1)(x-7)=2x^2-14x-x+7=2x^2-15x+7\ \ \Rightarrow\ \ k=-15\\\\3)\\(2x+7)(x+1)=2x^2+2x+7x+7=2x^2+9x+7\ \ \Rightarrow\ \ k=9\\\\4)\\(2x-7)(x-1)=2x^2-2x-7x+7=2x^2-9x+7\ \ \Rightarrow\ \ k=-9\\\\\\Ans. \ there\ are\ four\ solutions:\ k\in \{15;-15;9;-9\} \)


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