How do you divide $(6x^2-31x+5) / (x-5)$ ?

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Answer 1 · 2016-11-16T08:26:02

The answer is $=(6x-1)$

Let $f(x)=6x^2-31x+5$

Then, $f(5)=6*25-31*5+5=150-155+5=0$

The remainder is $0$ , so $f(x)$ is divisible by $(x-5)$

Let's do a long division

$color(white)(aaaa)$ $6x^2-31x+5$ $color(white)(aaaa)$ $∣$ $x-5$

$color(white)(aaaa)$ $6x^2-30x$ $color(white)(aaaaaaaa)$ $∣$ $6x-1$

$color(white)(aaaaaa)$ $0-x+5$

$color(white)(aaaaaaaa)$ $-x+5$

$color(white)(aaaaaaaa)$ $-0+0$

Now, we can factorise the numerator

$6x^2-31x+5=(6x-1)(x-5)$

Then,

$(6x^2-31x+5)/(x-5)=((6x-1)cancel(x-5))/cancel(x-5)=(6x-1)$

How do you divide (6x^2-31x+5) / (x-5)(6x^2-31x+5) / (x-5)?