3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)=
(x+1)(x+3)(3x-1)(x+2)
CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x).
Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division:
-1 | 3 17..27...7..-6
......3 -3 -14 -13...6
......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero.
The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor:
-3 | 3 14..13..-6
......3..-9 -15...6
.....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above).
6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)=
(x-1)(x+3)(3x-1)(2x-1)
CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor.
1 | 6...7 -27..17..-3
.....6...6..13 -14...3
.....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor.
-3 | 6..13 -14...3
......6 -18..15..-3
......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).
By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them:
HCF=(x+3)(3x-1)=3x^2+8x-3