Which of the following binomials is a factor of the expression 8y 2 – 10y – 3?

Question

F. (4y – 1)

G. (4y + 1)

H. (2y + 3)

J. (y – 3)

Answer 1

8y² -10y -3

= 8y² - 12y +2y -3

= 4y( 2y -3) +1 (2y -3)

= (4y +1) ( 2y -3)

Factors are ( 4y +1) and (2y -3)

G is correct

Answer 2

Divide the polynomial by each of the binomials. If there is a remainder, then the binomial is not a factor. If there is no remainder, the binomial is a factor. I'll do the first one, 4y - 1:

2y - 2

4y-1 ) 8y² - 10y - 3

-(8y² - 2y)

---------------

-8y - 3

-(-8y + 2)

-------------

-5

No this binomial (4y-1) has a remainder of -5, so it is not a factor. You try the other three.

Answer 3

One approach is to factor the polynomial:

The basic forms would be

(8y +/- 3 )(y +/- 1 )

(8y +/- 1)(y +/- 3 )

(4y +/- 3 )(2y +/- 1 )

(4y +/- 1 )(2y +/- 3 )

Trial and error shows that

(4y+1)(2y-3) is the factorization, so G.

(NOTE: there's nothing that says the factorization must have 'nice' integer coefficients, so if nothing worked in the above trial/error tests, it does not neccesarily follow that the polynomial can't be factored. If you've learned the quadratic formula there's a way to check whether a quadratic can be factored or not)

Another approach is to remember that of the given factors, their roots must also be roots of the original polynomial.

F) 4y-1 = 0 implies y=1/4

G) 4y+1=0 implies y=-1/4

H) 2y+3 = 0 implies y=-3/2

J) y-3 = 0 implies y=3

Plug each of these into the original polynomial, 8y^2-10y-3 shows that only y=-1/4 is a root, hence G)

There are other approaches, but these two are the quickest that came to mind