How do you factor $x^{3} - 12 x^{2} + 41 x - 42$ $x^3-12x^2+41x-42$ ?

Question

How do you factor $x^{3} - 12 x^{2} + 41 x - 42$ $x^3-12x^2+41x-42$ ?

1 Answer

Impact of this question

2451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T16:25:12

$(x - 2) (x - 3) (x - 7)$ $(x-2)(x-3)(x-7)$

Explanation:

If the roots of the cubic are integer, then they must divide into 42. Possible candidates, plus or minus: 1, 2, 3, 6, 7, 14, 21, 42.

A little exploration with a calculator reveals that both $x = 2$ $x=2$ and $x = 3$ $x=3$ are roots, so we can immediately factor the expression as
$(x - 2) (x - 3) (x - 7)$ $(x-2)(x-3)(x-7)$ , deducing from $42 = 2 \cdot 3 \cdot 7$ $42=2*3*7$ that the third root must be 7.

Science

Math

Humanities

... and beyond

How do you factor $x^{3} - 12 x^{2} + 41 x - 42$ $x^3-12x^2+41x-42$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor x3−12x2+41x−42x^3-12x^2+41x-42?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $x^{3} - 12 x^{2} + 41 x - 42$ $x^3-12x^2+41x-42$ ?