really confuzzled by this question?

Question

ok, here it is
Write the function as a product of linear and irreducible quadratic factors, all with real coefficients.
f(x) = 2x³+7x²+6x+9

any help is appreciated , thanx a lot~!! ^-^

Answer 1

The possible factors are going to come from the factors of the last coefficient (9) divided by the factors of the first coefficient (2).

So that would be {1, 3, 9, 1/2, 3/2, 9/2, -1, -3, -9, -1/2, -3/2, -9/2}

Positive roots aren't going to work because everything is added. Try some numbers until you get a zero.
f(-1) = -2 + 7 - 6 + 9 --> 8 (no good)
f(-3) = 2(-27) + 7(9) + 6(-3) + 9 --> -54 + 63 + -18 + 9 --> 0, good!
So (x + 3) will be a factor.

Now just use synthetic division:
...................... 2x² + x + 3
x + 3 ) 2x^3 + 7x² + 6x + 9
........... 2x^3 + 6x²
....................... x² + 6x
....................... x² + 3x
.............................. 3x + 9
.............................. 3x + 9

So you have (x + 3)(2x² + x + 3)

Now you can use the quadratic formula on the last part:
a = 2, b = 1, c = 3

..... -1 ± sqrt( 1² - 4(2)(3) )
x = -------------------------------
...................... 4

x = [-1 ± sqrt(-23) ] / 4

Oh, that won't reduce to integer or real roots...

So just leave it as:
f(x) = (x + 3)(2x² + x + 3)

Answer 2

Breaks down into two sums; set up the x's and trial and error
knowing the structure of the outcome.

( x^2 +2 x +3 )( 2 x + 3 )

x^3 x^2 x
x^2 x
______________
2x^3+ 7x^2 + 6x + 9

This way gives them all
( 2 x^2 + x + 3 )( x + 3 )

x^3 x^2 x
x^2 x
______________
2x^3+ 7x^2 + 6x + 9

Answer 3

2x³+7x²+6x+9
=(x+3)(2x^2+x+3)
//dividing f(x) by (x+3) since f(-3)=0