Math Help? What is the factored form of the​ polynomial? x^4+8x^3-35x^2-294x?

Hello,

x⁴ + 8x³ - 35x² - 294x =

let's first factor out x:

x (x³ + 8x² - 35x - 294) =

let's subtract and add x in parentheses:

x (x³ + 8x² - 35x - x + x - 294) =

x (x³ + 8x² - 36x + x - 294) =

let's group terms as:

x [(x³ - 36x) + (8x² + x - 294)] =

let's factor x out of the binomial in brackets:

x [x (x² - 36) + (8x² + x - 294)] =

let's further factor x² - 36 as a difference of squares:

x [x (x² - 6²) + (8x² + x - 294)] =

x [x (x + 6)(x - 6) + (8x² + x - 294)] =

as for the trinomial in brackets, let's subtract and add 48x:

x [x (x + 6)(x - 6) + (8x² - 48x + 48x + x - 294)] =

x [x (x + 6)(x - 6) + (8x² - 48x + 49x - 294)] =

let's group terms as:

x {x (x + 6)(x - 6) + [(8x² - 48x) + (49x - 294)]} =

let's factor 8x out of the first binomial and 49 out of the second one in brackets, yielding the common factor (x - 6):

x {x (x + 6)(x - 6) + [8x (x - 6) + 49(x - 6)]} =

let's factor out (x - 6):

x {x (x + 6)(x - 6) + [(x - 6)(8x + 49)]} =

let's factor out (x - 6) further:

x {(x - 6) [x (x + 6) + (8x + 49)]} =

x (x - 6) (x² + 6x + 8x + 49) =

x (x - 6) (x² + 14x + 49) =

let's note that the remaining trinomial is a perfect square:

x (x - 6) [x² + (2 ∙ x ∙ 7) + 7²] =

then let's apply the factoring rule a² + 2ab + b² = (a + b)², ending with:


x (x - 6)(x + 7)²



I hope it's helpful

Let f(x)=x^4+8x^3-35x^2-294x=0, then
f(x)=x(x^3+8x-35x-294)=0
=>x=0 is a real root
x^3+8x^2-35x-294=0
=>x=6, x= -7 (a repeated root)
=>f(x)=x(x-6)(x+7)^2

Check for the roots by division :
1 + 8 - 35 - 294 | 6
...+ 6+ 84+ 294 |
----------------------|
1 +14+49 |.........| -7
....- 7..-49 |
--------------|
1 + 7 |.......| -7
.....-7 |
--------|
1
=>
x^3+8x^2-35x-294=
(x-6)(x+7)^2
=>
f(x)=x(x-6)(x+7)^2

(x)(x+7)(x+7)(x-6) is the factored form of the polynomial.