Can anyone factor these problems?

Question

I've tried over and over but I just can't get it.
The first one is x^2+3x+1.
The second one is x^6-2x^5+7x^4.

Answer 1

x^2 + 3x + 1

Can't be factored further, without using the quadratic. It factors to:

(x + 3/2 - sqrt(5)/2)(x + 3/2 + sqrt(5)/2)

That's messier than the original, in my opinion, and probably not what your teacher wants to see.

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x^6 - 2x^5 + 7x^4
x^4 (x^2 - 2x + 7)

Can't be factored further. The discriminant of the quadratic is 4-28 = -24, so it would require imaginary numbers to factor the quadratic.

Answer 2

1.x^2+3x+1=(x-a)(x-b)., where a&b are zeros of x^2+3x+1
which are -3+/-sqrt(9-4)]/2=1/2[-3+/-sqrt(5)].thes are irrational roots
2.x^6-2x^5+7x^4=x^4(2x^2-2x+7).ANS