if (x-7) is a factor of 2x^2-11x+k what is the value of k

if (x-7) is a factor of \(2x^2-11x+k\) what is the value of k

I literally have no idea what to do here.

Because a factor is x-7, therefore one root of the equation of \(2x^2-11x+k=0\), must be 7.

The roots are:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x_{1,2} &=& \frac{ b \pm \sqrt{b^2-4ac} } { 2a } \\ \hline \end{array}\)

So we have:

\(\begin{array}{|rcll|} \hline 2x^2 -11x + k &=& 0 \qquad \qquad a = 2 \qquad b=-11 \qquad c = k \\\\ x_{1,2} &=& \frac{ 11 \pm \sqrt{11^2-4\cdot 2 \cdot k} } { 2\cdot 2 } \\\\ \mathbf{x_{1,2}} &\mathbf{=}& \mathbf{\frac{ 11 \pm \sqrt{121-8k} } { 4 } }\\ \hline \end{array}\)

We can to equal:
\(\begin{array}{|rcll|} \hline \mathbf{x_\text{root}} &\mathbf{=}& \mathbf{\frac{ 11 \pm \sqrt{121-8k} } { 4 } } \qquad x_\text{root} = 7\\\\ \frac{ 11 \pm \sqrt{121-8k} } { 4 } &=& 7 \qquad & | \qquad \cdot 4 \\ 11 \pm \sqrt{121-8k} &=& 7 \cdot 4 \\ 11 \pm \sqrt{121-8k} &=& 28 \qquad & | \qquad -11 \\ \pm \sqrt{121-8k} &=& 28 -11 \\ \pm \sqrt{121-8k} &=& 17 \qquad & | \qquad \text{square both sides} \\ 121-8k &=& 17^2 \\ 121-8k &=& 289 \qquad & | \qquad \cdot (-1) \\ 8k &=& 121-289 \qquad & | \qquad +121 \\ 8k &=& -168 \qquad & | \qquad :8 \\ \mathbf{k} &\mathbf{=}& \mathbf{-21} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2x^2 -11x -21 &=& (x-7)\cdot(2x+3) \\ \hline \end{array}\)

the value of k is -21

If (x-7)  is  a factor, this implies that  x = 7   is a root......so we have that

2(7)^2  - 11(7)   +   k  =  0        simplify

98 - 77  + k  = 0

21  + k  = 0     subtact 21 from both sides

k = -21

See the graph here to confirm this: https://www.desmos.com/calculator/xmg3aleqko