im stuck on factoring (x-1)(2x^3+13x^2+8x-35)
do i use synthetic division? if so what zero?
thnks.
To factor the expression (x-1)(2x^3+13x^2+8x-35), you do not need to use synthetic division. In this case, you can use the factor theorem or synthetic substitution to find the zero that will help you factor the expression.
The factor theorem states that if a polynomial function is divided by (x-a) and the remainder is zero, then (x-a) is a factor of the polynomial.
To find the zero, you can set the expression inside the first set of parentheses, (x-1), to zero and solve for x:
x - 1 = 0
x = 1
So, the zero is x = 1.
Now, let's use this zero to factor the expression:
Substitute x = 1 into the expression (2x^3+13x^2+8x-35):
(2(1)^3+13(1)^2+8(1)-35)
= 2 + 13 + 8 - 35
= -12
Now, divide the expression (2x^3+13x^2+8x-35) by (x-1) using synthetic division or long division:
2 13 8 -35
----------------------
1 | 2 13 8 -35
- 2 15 23
-------------
2 15 23 -12
The resulting quotient is 2x^2 + 15x + 23 and the remainder is -12.
Therefore, you can rewrite the original expression as (x-1)(2x^2 + 15x + 23).