Question

factorize x⁴-2x³-7x²+8x+12
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Answer 1

Step-by-step explanation:

How do I factorize x4−2x3−7x2+8x+12 ?

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By trial and error the roots are factors of 12 that is -12,-6,-4,-3,-2,-1,0,1,2,3,4,6,12.

Let f(x)=x⁴-2x³-7x²+8x+12

f(-1)=(-1)⁴-2(-1)³-7(-1)²+8(-1)+12=0

So (x+1) is a factor.

Using synthetic division

-1| 1 -2 -7 8 12

| 0 -1 3 4 -12

|1. -3 -4 12 0*

Where 0* is a zero only but it indicates the remainder of the division.

So the polynomial is written as

x⁴-2x³-7x²+8x+12=(x+1)(x³-3x²-4x+12)

Now consider,

g(x)=x³-3x²-4x+12

similarly if we test as above

g(-2)=(-2)³-3(-2)²-4(-2)+12=-8–12+8+12=0

Now again computing synthetic division

-2| 1 -3 -4. 12

|0 -2 10 -12

|1 -5 6 0

So the g(x)=(x+2)(x²-5x+6)

x²-5x+6=x²-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)

g(x)=(x+2)(x-2)(x-3)

f(x)=x⁴-2x³-7x²+8x+12

f(x)=(x+1)g(x)

f(x)=(x+1)(x+2)(x-2)(x-3).

So roots are x=1 ,2,-2,3.

Answer 2

Step-by-step explanation:

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