376. ( x + 4) 2 ( x – 6)
There is only one possible positive real root, so try for a negative root first.
Because –4 is a root, (x + 4) is a factor. Writing the coefficients in the bottom row in the corresponding trinomial, you have
x 3 + 2x 2 – 32x – 96 = (x + 4)(x 2 – 2x – 24)
Now factor the quadratic trinomial.
= (x + 4)(x + 4)(x – 6)
The solution x = –4 is a double root, so you can write the factorization as (x + 4)2(x – 6).
377. ( x – 1)( x + 2)( x + 5)( x – 3)
There are two or no possible positive real roots, so there’s no advantage to trying one type of root or the other. But, trying a positive number first, you have
Now, using the coefficients in the bottom row, try a negative number (because you now have two negative and one positive remaining):
Because 1 and –2 are roots, (x – 1) and (x + 2) are factors. Writing the coefficients in the bottom row in the corresponding trinomial, you have
x 4 + 3x 3 – 15x 2 – 19x + 30
= (x – 1)(x + 2)(x 2 + 2x – 15)
Now factor the quadratic trinomial.
= (x – 1)(x + 2)(x + 5)(x – 3)
378. ( x + 2)( x + 3)( x – 1)( x + 4)
There is only one possible positive real root, so try for a negative root first.
Now, ...
