# Express x^4-x^3-5x^2-x-6 in the form (x^2+a)(x+b)(x+c), where a, b and c are constants to be determined. Thanks!

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`x^4-x^3-5x^2-x-6 = (x^2+a)(x+b)(x+c)`

`x^4-x^3-5x^2-x-6 = (x^2+a)(x^2+(b+c)x+bc)`

`x^4-x^3-5x^2-x-6 =

x^2(x^2+(b+c)x+bc)+a(x^2+(b+c)x+bc)`

`x^4-x^3-5x^2-x-6 =

x^4+(b+c)x^3+bcx^2+ax^2+a(b+c)x+abc`

`x^4-x^3-5x^2-x-6 =

x^4+(b+c)x^3+(a+bc)x^2+(ab+ac)x+abc`

Now let us compare the components;

x^3 `rarr -1 = b+c` -----(1)

x^2 `rarr -5 = a+bc` -----(2)

x     `rarr -1 = ab+ac` -----(3)

con. `rarr -6 = abc` ----(4)

Now we can see that `(1)xxa = (3)`

`-1xxa = ab+ac `

`-1 = ab+ac`

Therefore;

`-1a = -1`

`a = 1`

`(1)xxb-(2)`

`-b-(-5) = b^2+bc-a-bc`

`5-b = b^2-1`

`b^2+b-6 = 0`

`(b+3)(b-2) = 0 `    (you will get this by factoring)

`b = -3` or `b = 2`

When `b = -3`

`b+c = -1`

`c = -1+3`

`c = 2`

When `b = 2`

`b+c = -1`

` c = -1-2`

` c = -3`

So the factors would be;

`x^4-x^3-5x^2-x-6 = (x^2+1)(x+2)(x-3)`

OR

`x^4-x^3-5x^2-x-6 = (x^2+1)(x-3)(x+2)`

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Apply remainder Theorem ,we find zero this function is -2,3

`x^4-x^3-5x^2-x-6`

`=x^4+2x^3-3x^3-6x^2+x^2+2x-3x-6`

`=(x+2)(x^3-3x^2+x-3)`

`=(x+2)(x-3)(x^2+1)`

`=(x^2+1)(x+2)(x-3)`

`a=1,b=2, c=-3`

``