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find the value of p and of q for which x-3 is a common factor of the expressions [tex]x^{2} + (p + q)x-q[/tex] and [tex]2x^2 + (p-1)x + (p+2q)[/tex]

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Step-by-step explanation:

Use the Factor Theorem,

" if x-a is a factor of f(x), then f(a) =0,

So here, since x-3 is a factor then f(3)=0,

So first step, plug in 3 for x for the serperate equations.

[tex] {3}^{2} + ( p + q)3 - q = 0[/tex]

[tex]2(3) {}^{2} + (p - 1)3 + (p + 2q) = 0[/tex]

Simplify both equations,

[tex]9 + 3p + 2q = 0[/tex]

[tex]15 + 4 p + 2q = 0[/tex]

Isolate the constants,

[tex]3p + 2q = - 9[/tex]

[tex]4p + 2q = - 15[/tex]

We have a system of equations so let eliminate a variable, by subtracting the two equations.

[tex] - p = 6[/tex]

[tex]p = - 6[/tex]

Plug p back in for any one of the equations to find q.

[tex]3( - 6) + 2q = - 9[/tex]

[tex] - 18 + 2q = - 9[/tex]

[tex]2q = 4.5[/tex]

So p is -6

q is 4.5

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