factorize(因式分解): (x+1)(x+3)(x+4)(x+6)-280?

Question

how?

Answer 1

Expand:
x^4 + 14x^3 + 67x^2 + 126x + 72 - 280
= x^4 + 14x^3 + 67x^2 + 126x - 208
©
By synthetic division

1 ... 14 ... 67 .... 126 .... -208 ....|| 1
....... .1 .... 15 ..... 82 ....... 208
------------------------------------------
1 ... 15 ... 82 .... 208 .......... 0 .. (x-1) is a factor

1 ... 15 .... 82 .... 208 .. || (-8)
....... -8 ... -56 ... -208
--------------------------------
1 .... 7 ..... 26...... 0 ....... (x+8) is another factor

Thus (x+1)(x+3)(x+4)(x+6)-280
= (x-1)(x+8) (x^2 + 7x + 26)

Answer 2

First multiply everything
(x+1)(x+3)(x+4)(x+6)-280 =
(x^2+4x+3)(x^2+10x+24) -280 =
x^4 +14x^3 +67x^2 +126x - 208

Now think about the factors of -208:
1, 2, 2, 2, 2, 13 (any could be multiplied together, so it could be 1, 2, 4, 26)
An odd number of the factors will have to be negative

Now try dividing by (x-1) or (x+1).
(x-1) works, leaving
x^3 +15x^2 +82x +208

Now, you need three factors and either one or three are positive and zero or two are negative.

Answer 3

(x+1)(x+3)(x+4)(x+6) - 280

Try multiplying it through first...
(x + 1)(x + 3) = x^2 + 4x + 3
(x + 4)(x + 6) = x^2 + 10x + 24

(x^2 + 4x + 3)(x^2 + 10x + 24)
= x^4 + 10x^3 + 24x^2 + 4x^3 + 40x^2 + 96x + 3x^2 + 30x + 72
= x^4 + 14x^3 + 67x^2 + 126x + 72

So... now you have:
x^4 + 14x^3 + 67x^2 + 126x + 72 - 280
= x^4 + 14x^3 + 67x^2 + 126x - 208
= (x + 8)(x^3 + 6x^2 + 19x - 26)
= (x + 8)(x - 1)(x^2 + 7x + 26)

That doesn't factor any further.

Answer 4

If x+7/2=x+3,5=t

-->(x+1)(x+3)(x+4)(x+6)-280 =
(t-2,5)·(t-0,5)·(t+0,5)·(t+2,5) -280 =
(t^2-6,25)·(t^2-0,25) -280 =
t^4-6,5 t^2 - 278,375 =0 <-->

t^2 = 81/4 or -55/4 -->

t= ±9/2 or ±√55/2 ·i

x= 1, -8 , (-7±√55 i)/2

--> (x+1)(x+3)(x+4)(x+6)-280 =
(x-1)·(x+8) ·[x-(-7+√55 i)/2] · [x-(-7-√55 i)/2 or
(x-1)·(x+8)· (x^2+7x+26)

Saludos.

Answer 5

(x+1)(x+3)(x+4)(x+6)-280

1:
(x+1) x ( x+3) = x^2 + 4x + 3

2.
(x^2 + 4x +3) x (x +4) = x^3 + 8x^2 + 11x +
12
3.
(x^3 + 8x^2 + 11x +12) x (x+6)
= X^4 + 14x^3 + 59^2 + 78x + 72

4.
(x+1)(x+3)(x+)(x+6) -280
= x^4 + 14x^3 + 59x^2 + 78x +72 -280
=x^4 + 14x^3 + 59x^2 + 78X -208

ANSWER: x^4 + 14x^3 + 59x^2 + 78X -208

Answer 6

first let us rearrange the terms.
1+6=3+4
so i'll write (x+1)(x+6)as one group and (x+3)(X+4) as another.

a)
(x+1)(x+6)=x^2+6x+x+6
=x^2+7x+6

b)
(x+3)(x+4)=x^2+3x+4x+12
=x^2+7x+12

c)
multiplying both gives us (x^2+7x+6)(x^2+7x+12)
=x^4+7x^3+12x^2+7x^3+49x^2+84x+6x^2+42x+72)
=x^4+14x^3+67x^2+126x+72
d)
now subtract 280 from x^4+14x^3+67x^2+126x+72

e)
=x^4+14x^3+67x^2+126x-208

Answer 7

you can see: 1+6=3+4=7
(x+1)(x+6)(x+3)(x+4)-280=(x^2+7x+6)(x^2+7x+12)-280
t=x^2+7x+6
we have:
t(t+6)-280=t^2+6t-280=(t-14)(t+20)=(x^2+7x-8)(x^2+7x+26)

Answer 8

(x+1)(x+3)(x+4)(x+6)-280
(x+1)(x+6)(x+3)(x+4)-280
(x^2+7x+6)(x^2+7x+12)-280

Put x^2+7x=A

(A+6)(A+12)-280
A^2+18A+72-280
A^2+18A-208
A^2+26A-8A-208
A(A+26)-8(A+26)
(A-8)(A+26)

put back A=x^2+7x

(x^2+7x-8)(x^2+7x+26)

x^2+7x-8 can further be factorised

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

So the answer is

(x-1)(x+8)(x^2+7x+26)