Oxford PAT 2012, Question 5

2012_page5

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2 thoughts on “Oxford PAT 2012, Question 5

  1. Thank you so much for these solutions, it’s helping me study so much!
    I was wondering on the first part of this problem, where it wants us to “show” that x=1 is a solution… I know this means that (x-1) must be a factor, so I did polynomial long division to show that, but it took me a long time. On the contrary, I see you were able to jump right to (x-1)(x^2 – 5x – 14). Is there a way to find this factored version right away, or did you have other work that you just didn’t put on here?
    Thanks so much again!

    1. It’s a pleasure, I’m glad it’s helping.

      In this case you can do the factorization fairly easily because, if (x-1) is a factor, you can see that the other factor must be (x² + Ax + B). By looking at the constant part you can see that (-1)*(B)=14, so B = -14. By looking at coefficients of x² you can see that (-1x²)+(Ax²)=-6x², so A = -5. You can then look at the coefficient of x just to check that (x-1) is a factor.

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