Random Sec 4 student's answer to Shreyas's Secondary 4 A Maths Singapore question.
Actually this solution is correct.
For (a), we come up with 2 simultaneous equations in terms of “a” and “b”, using factor theorem and remainder theorem.
Then, solve the 2 simultaneous equations using substitution (as per the photo) or elimination.
For (b), we find the first factor using calculator. Usually, we’ll pick the one with an integer. For instance, x=-3.
However, when we present the working, we write:
f(-3)= 4(-3)³ + 8(-3)² - 15(-3) - 9 = 0
After that, we use long division or synthetic method to get a quadratic expression, which is then factorised into 2 factors.
For (a), we come up with 2 simultaneous equations in terms of “a” and “b”, using factor theorem and remainder theorem.
Then, solve the 2 simultaneous equations using substitution (as per the photo) or elimination.
For (b), we find the first factor using calculator. Usually, we’ll pick the one with an integer. For instance, x=-3.
However, when we present the working, we write:
f(-3)= 4(-3)³ + 8(-3)² - 15(-3) - 9 = 0
After that, we use long division or synthetic method to get a quadratic expression, which is then factorised into 2 factors.
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