im stuck on factoring (x-1)(2x^3+13x^2+8x-35)
do i use synthetic division? if so what zero?
thnks.
synthetic division would be an easy way to check for roots p/q where p divides 35 and q divides 2.
Check for whole roots first: 1,5,7,35
A little trial shows that
2x^3+13x^2+8x-35 = (x+5)(2x^2+3x-7)
To factor the expression (x-1)(2x^3+13x^2+8x-35), you do not need to use synthetic division. Synthetic division is used to divide a polynomial by a binomial of the form (x-a), where "a" is a constant. In this case, you have a product of two binomials.
To factor this expression, you can use the method of factoring by grouping. Here's how to do it:
Step 1: Recognize that (x-1) is a common factor in each term of the expression.
(x-1)(2x^3+13x^2+8x-35)
= (x-1) [ 2x^3 + 13x^2 + 8x - 35 ]
Step 2: Group the terms inside the brackets:
= (x-1) [ (2x^3 + 8x) + (13x^2 - 35) ]
Step 3: Factor out the common terms from each group:
= (x-1) [ 2x(x^2 + 4) + 13(x^2 - 5) ]
Step 4: Factor out any common factors again if possible:
= (x-1) [ 2x(x^2 + 4) + 13(x^2 - 5) ]
= (x-1) [ 2x(x^2 + 4) + 13(x^2 - 5) ]
= (x-1) [ 2x(x^2 + 4) + 13(x^2 - 5) ]
Now the expression is in factored form. The next step would be to check for any further factoring you can do within the brackets (if possible) or to evaluate the equation as it is.