How do you determine the binomial factors of $x^3-x^2-49x+49$ ?

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Answer 1 · 2016-12-04T08:38:57

The binomial factors of $x^3-x^2-49x+49$ are:

$(x^2-49)$ , $(x-1)$ , $(x-7)$ , $(x+7)$

The given cubic factors by grouping:

$x^3-x^2-49x+49 = (x^3-x^2)-(49x-49)$

$color(white)(x^3-x^2-49x+49) = x^2(x-1)-49(x-1)$

$color(white)(x^3-x^2-49x+49) = (x^2-49)(x-1)$

Note that both $color(blue)((x^2-49))$ and $color(blue)((x-1))$ are binomial factors.

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

Hence we find:

$x^2-49 = x^2-7^2 = (x-7)(x+7)$

So $color(blue)((x-7))$ and $color(blue)((x+7))$ are also binomial factors.

Multiplying either of these by $(x-1)$ will result in a trinomial, so there are no other binomial factors:

$(x-7)(x-1) = x^2-8x+7$

$(x+7)(x-1) = x^2+6x-7$

The complete list of polynomial factors of $x^3-x^2-49x+49$ in descending degree is:

$x^3-x^2-49x+49$

$x^2-49$

$x^2-8x+7$

$x^2+6-7$

$x-7$

$x+7$

$x-1$

$1$

How do you determine the binomial factors of x^3-x^2-49x+49x^3-x^2-49x+49?