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

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Answer 1 · 2016-11-16T08:26:02

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

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

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

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

Let's do a long division

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

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

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

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

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

Now, we can factorise the numerator

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

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)6x2−31x+5x−5(6x^2-31x+5) / (x-5)?