Question

How to factor the polynomial `P(x)` and then solve the equation `P(x)=0` for the given below? (1) `P(x)=x^3-6x^2-x+6` (2) `P(x)=x^4-x^3-19x^2+49x-30`

Answer 1

`"see explanation"`

Explanation:

`(1)`

`color(blue)"factor by grouping"`

`P(x)=color(red)(x^2)(x-6)color(red)(-1)(x-6)`

`"take out the common factor "(x-6)`

`P(x)=(x-6)(color(red)(x^2-1))larr"difference of squares"`

`color(white)(P(x))=(x-6)(x-1)(x+1)`

`P(x)=0`

`(x-6)(x-1)(x+1)=0`

`"equate each factor to zero and solve for "x`

`x-6=0rArrx=6`

`x-1=0rArrx=1`

`x+1=0rArrx=-1`

`(2)`

`"note that "`

`P(1)=1-1-19+49-30=0`

`rArr(x-1)" is a factor"`

`"dividing "x^4-x^3-19x^2+49x-30" by "(x-1)" gives"`

`P(x)=(x-1)(x^3-19x+30)`

`"note that "2^3-19(2)+30=8-38+30=0`

`rArr(x-2)" is a factor"`

`"dividing "x^3-19x+30" by "(x-2)" gives"`

`P(x)=(x-1)(x-2)(x^2+2x-15)`

`color(white)(P(x))=(x-1)(x-2)(x+5)(x-3)`

`P(x)=0`

`(x-1)(x-2)(x+5)(x-3)=0`

`x-1=0rArrx=1`

`x-2=0rArrx=2`

`x+5=0rArrx=-5`

`x-3=0rArrx=3`