If $p(x) = x^2 -1$ and $q(x) = 5(x-1)$ , what is $(p-q)(x)$ ?

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Answer 1 · 2017-12-13T00:30:06

$x^2-5x+4$

$(p-q) (x)=p(x)-q(x)=(x^2-1)-5*(x-1)=x^2-5x+4$

Answer 2 · 2017-12-13T00:34:30

$x^2- 5x+4$

Question, $(p-q)(x)$

Here both $p$ and $q$ are functions assuming $(p-q)(x)$ means $p(x)-q(x)$ .

$(p-q)(x) =$
$p(x)-q(x)$

$= p(x) - q(x)$
$= (x^2-1) - {5(x-1)}$
$= x^2-1 - {5x-5}$
$= x^2-1 - 5x+5$
$= x^2- 5x+4$

Thus, $(p-q)(x)=x^2- 5x+4$ .

If p(x) = x^2 -1p(x) = x^2 -1 and q(x) = 5(x-1)q(x) = 5(x-1), what is (p-q)(x)(p-q)(x)?