Download There will be a few problems on the provincial exam in which

There will be a few problems on the provincial exam in which Students are asked to analyze
a polynomial expression. These problems tend to be quite challenging. In order to solve these
problems students will need to have a solid understanding of how to factor the following types of
polynomials.
• Simple Trinomials like: x2 + 7x + 10 = (x + 5)(x + 2)
• Complex Trinomials with a leading coefficient like: 3x2 + 17x + 10 = (3x + 2)(x + 5)
• Perfect Square Trinomials: These are of the form (x+y)2 = (x+y)(x+y) = x2 +2xy +y 2 ,
Example: x2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)2
• Difference of Squares: These are polynomials of the form x2 −y 2 = (x+y)(x−y), Example:
4x2 − 9 = (2x + 3)(2x − 3)
Students will be asked to use their understanding of the factoring polynomials to solve the following
problems:
1. Katie simplified the expression (x + b)(x + c), where b < 0 and c < 0, to the form x2 + gx + k.
What must be true about g and k?
(a) g < 0 and k > 0
(b) g < 0 and k < 0
(c) g > 0 and k > 0
(d) g > 0 and k < 0
Fist lets explore what we know. We know that since b and c are less then 0 they must be
negative numbers. Also we know that Katie simplified the expression. SO I am going to do
that first. To simplify this expression I will simply multiply the two monomials together.
(x + b)(x + c)
x2 + bx + cx + bc
x2 + (b + c)x + bc
Now lets compare these two expressions:
x2 + (b + c)x + bc and x2 + gx + k
We see in comparing these that (b + c) corresponds to g and bc corresponds to k.
Now since b and c are less then zero (aka negative numbers) we can determine some things
about g and k. If two negative numbers are added together, (b + c), they give another
negative number. So we know that g < 0 or g is a negative number. If two negative numbers
are multiplied, bc, we know the answers is a positive number. Therefore k > 0 or k is a
positive number. The answer to this question is a.
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2. How many integer values are there for k for which 4x2 + kxy − 9y 2 is factorable?
First off for a quadratic polynomial to be factorable that means we can factor it down into
two parts. x2 + 4x + 4 −→ (x + 2)(x + 2).
We need to take 4x2 + kxy − 9y 2 and factor it into (?x+?y)(?x+?y)
Knowing a bit about factoring I will make some educated guesses as to what this quadratic
could factor to.
Does 4x2 + kxy − 9y 2 = (2x + 3y)(2x − 3y) ?
I am going to guess that this configuration will factor since:
2x × 2x = 4x2 and 3y × −3y = −9y 2
If I then multiply my guess out I get:
(2x + 3y)(2x − 3y) =
4x2 − 6xy + 6xy − 9y 2 =
4x2 + 0xy − 9y 2
Therefore one possible value for k is 0.
Now we need to see if we can find others.
Does 4x2 + kxy − 9y 2 = (x + y)(4x − 9y) ?
I am going to guess that this configuration will work since:
1x × 4x = 4x2 and 1y × −9y = −9y 2
If I then multiply my guess out I get:
(x + y)(4x − 9y) =
2
4x − 9xy + 4xy − 9y 2 =
4x2 − 5xy − 9y 2
Therefore another possible value for k is -5.
Your task is to continue to use this process to find out how many values of k exist in which
4x2 + kxy − 9y 2 will factor. Hint: You only need to try all the combinations of the factors
of 4 and 9.
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