2015 methods exam 2 mcq 3

[sailinginwater]

Why is multiple choice 3 C and not A?

[S_R_K]

If the graph of a polynomial has an x-intercept at x = n, then (x – n) is a factor of the polynomial. This is true for any value of n.

For that question, since the x-intercepts of the graph are at x = b, x = c, x = d, then the polynomial has factors (x – b), (x – c), (x – d).

Many students incorrectly chose option A because they reasoned that if b is a negative number, then (x + b) is a factor (thinking that you need to "flip the sign"). This is not correct. If you substitute in x = b into the factor, you should get 0. But if you substitute x = b into (x + b), you'll get 2b, not 0. Hence (x + b) is not a factor.

[sailinginwater]

If the graph of a polynomial has an x-intercept at x = n, then (x – n) is a factor of the polynomial. This is true for any value of n.

For that question, since the x-intercepts of the graph are at x = b, x = c, x = d, then the polynomial has factors (x – b), (x – c), (x – d).

Many students incorrectly chose option A because they reasoned that if b is a negative number, then (x + b) is a factor (thinking that you need to "flip the sign"). This is not correct. If you substitute in x = b into the factor, you should get 0. But if you substitute x = b into (x + b), you'll get 2b, not 0. Hence (x + b) is not a factor.

Thanks 
I just realised that a lot of the time the incorrect alternatives in multiple choice questions are common mistakes made?

[Orb]

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