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Answer:
[tex]k=-\dfrac{11}{2}[/tex]
[tex]\left(x+\dfrac{7}{2}\right)[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{5.7 cm}\underline{Factor Theorem}\\\\If f$(x)$ is a polynomial, and f$(a) = 0$,\\then $(x-a)$ is a factor of f$(x)$.\end{minipage}}[/tex]
Given polynomial function:
[tex]p(x)=x^2-kx+7[/tex]
Apply the Factor Theorem:
[tex]\begin{aligned} \textsf{If $(x + 2)$ is a factor of $p(x)$ then: \quad}p(-2)&=0\\\\\implies (-2)^2-k(-2)+7&=0\\4+2k+7&=0\\2k+11&=0\\2k&=-11\\k&=-\dfrac{11}{2}\end{aligned}[/tex]
Substitute the found value of k into the original function:
[tex]\implies p(x)=x^2-\left(-\dfrac{11}{2}\right)x+7[/tex]
[tex]\implies p(x)=x^2+\dfrac{11}{2}x+7[/tex]
As (x + 2) is factor of the polynomial, and the leading coefficient of the function is 1, the other factor will be (x + q) where q is a constant to be found.
[tex]\begin{aligned}x^2+\dfrac{11}{2}x+7 & = (x+2)(x+q)\\& = x^2+qx+2x+2q\\& = x^2+(2+q)x+2q \end{aligned}[/tex]
Compare the constant of the given polynomial with the constant of the expanded factors:
[tex]\implies 2q=7[/tex]
[tex]\implies q=\dfrac{7}{2}[/tex]
Therefore:
[tex]\implies x^2+\dfrac{11}{2}x+7 = (x+2)\left(x+\dfrac{7}{2}\right)[/tex]
So the other factor of the given polynomial is:
[tex]\left(x+\dfrac{7}{2}\right)[/tex]