Question

Factorise? a) x^4+2x^3+3x^2+2x+1

Answer 1

`x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2`.

Explanation:

When you have a "symmetric polynomial", one where the coefficients are the same when you read them forwards and backwards, like `x^4+2x^3+3x^2+2x+1`, then it will have symmetric factors too.

Let us assume that the given fourth degree polynomial has a pair of symmetric, quadratic factors. The natural choice, since the first and last coefficients are both `1`, is:

`x^4+2x^3+3x^2+2x+1=(x^2+ax+1)(x^2+bx+1)`

Multiply the factors on the right:

`x^4+2x^3+3x^2+2x+1=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1`

So

`a+b=2`, thus `b=2-a`

`ab=a(2-1)=1`, thus `a^2-2a+1=0`

The only root for `a` is `a=1`, then also `b=1` and we end up with:

`x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2`.

Answer 2

`(x^2+x+1)^2`.

Explanation:

If we complete the square `x^4+2x^3`, we will get, `x^2` as the

the third term.

`:. x^4+2x^3+color(red)(3x^2)+2x+1`,

`=(x^4+2x^3+color(red)(x^2))+ul(color(red)(2x^2)+2x)+1`,

`={(x^2)^2+2(x^2)(x)+(x)^2}+2(x^2+x)+1`,

`=(x^2+x)^2+2(x^2+x)(1)+1^2`,

`=y^2+2y+1," where, "y=x^2+x`,

`=(y+1)^2`.

Sub.ing `y=x^2+x,` we find,

`"The Expression="(x^2+x+1)^2`.

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