How do you factor given that $f(6)=0$ and $f(x)=5x^3-27x^2-17x-6$ ?

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Answer 1 · 2016-11-11T17:02:42

We know that $(x-6)$ is a factor of $f(x)$ because of the given root.

Using synthetic division, I divided f(x) by (x-6) to get the other factor. enter image source here

$f(x)=(x-6)(5x^2+3x+1)$

Answer 2 · 2016-11-11T17:15:39

The only real factor of $f(x)$ are $(5x^2+3x+1)(x-6)$ . However, complex factors are $5(x-6)(x+(3+isqrt11)/10)(x+(3-isqrt11)/10)$

As $f(x)=5x^3-27x-17x-6$ and $f(6)=0$ , $(x-6)$ is a factor of $f(x)$ .

Hence $f(x)=5x^3-27x^2-17x-6$

$color(white)(XXXXXx)=5x^2(x-6)+3x(x-6)+1(x-6)$

$color(white)(XXXXXx)=(5x^2+3x+1)(x-6)$

As determinant (given by $b^2-4ac$ ) in $5x^2+3x+1$ is $3^2-4xx5xx1=-11<0$ , $5x^2+3x+1$ cannot be factorized further with real roots.

Hence only factor are $(5x^2+3x+1)(x-6)$ .

However for $5x^2+3x+1=0$ , $x=(-3+-sqrt(-11))/10$

Hence, we can have complex factors

$f(x)=5x^3-27x-17x-6$

= $5(x-6)(x+(3+isqrt11)/10)(x+(3-isqrt11)/10)$

How do you factor given that f(6)=0f(6)=0 and f(x)=5x^3-27x^2-17x-6f(x)=5x^3-27x^2-17x-6?