 # HEC Forum

Explore our forum below and start discussion!

### Challenges Wanted!

Here is your chance to challenge your tutor! Post your question and let them solve it. 8-)
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### Solve Me if You can!

For age 12+! Looking for challenge? Keep an eye on our challenge post and be the first one to solve it ... CORRECTLY!
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### Frequent Q & A

See what have been asked frequently from our students / parents
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New Posts
• ## G-E-O-ME-TRY

Another problem is on congruency of triangles to challenge the motivated students in plane geometry.
• ## G-E-O-ME-TRY

The problem below on Euclidean Geometry may pose a serious challenge to the mathematically gifted students based on common principles and theorems of plane geometry, in particular, congruency of two triangles.
• ## Yr10 James Ruse Exam Question

Hi, I thought I would share this. Found this challenging at first until I studied polynomials further for yr 10 course. Found this question in one of the James Ruse year 10 papers on the net. Part i) was a little confusing at first, I didn't see it in any text books. Found it finally in book 2 of the HEC Yr 10 Polynomials where it mentions under Polynomial Division, that the degree of the Remainder is always less than the degree of the Divisor. In this case the Divisor is quadratic and of degree 2. So the Remainder is linear of degree 1. Answer: The remainder always has a lower degree than the divisor. Since the divisor is quadratic R(x)=ax+b which is a general form of a linear expression (ie of degree 1). Part ii) Simply use the remainder theorem and sub. in 4 for P(x) and R(x). Then find 4a+b as common to both. Took a pic of the solution. Part iii) Use part ii) relationship for a and b variables, apply Remainder Theorem P(-1)=5 since divisor is (x-(-1)) or (x+1) to get answer. See pic of solution.

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