Patterns in Polynomials

Patterns can make math so much more easier, and interesting. They provide shortcuts that solves problems in half the time, and give you pointers to keep you in the right direction.

These three algebra tile models show a useful pattern that can be used to figure out which sign to use (negative or positive) when factoring polynomials. If it is completely shaded in, that means the whole expression is positive.

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If all the x tiles are shaded in on one side (it can be either horizontally or vertically), then that indicates there is a mix of positive and negative signs.

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If the model is checkered, with specifically the x squared tiles and ones tiles shaded in, then that means there are only negative signs.

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Another pattern which makes expanding polynomials easier is the one similar to FOIL. But instead of individually distributing the multiplication step-by-step, you can simply use basic math to expand.

(x - 3)(x - 7)

x\cdot x = x^2

-3\cdot -7 = 21

-3 + -7 = -10x

= x^2 - 10x + 21

You just multiply the first two terms together, and then the last two together. Then you add the last two terms in each bracket together.

This pattern always works on these simple expressions, and saves a lot of time, especially when solving word problems with many different steps in it. It can take a bit longer to write out the new expression after multiplying, and then circling or labelling to combine like terms. Patterns are useful!

Week 9 – Math 10

This week I learnt how to use an area model to expand an expression. It’s very useful since it organizes everything so you don’t miss a step while multiplying terms. Heres an example:

(x - 4)(x^2 - 6x + 3)

Draw a lowercase T and divide the bottom and right according to the number of terms in the expression. Since a binomial is being multiplied by a trinomial, the top is divided in two and the sides are divided in three.

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Now plug the numbers into their corresponding spots:

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Then, you simply multiply the numbers together! You can almost think of it like playing battleship:

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I have colour coded all the like terms together. The ones that are the same colour are added together, and the final answer is x^3 - 10x^2 + 27x - 12 .

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Week 8 – Math 10

This week, I got half of the question correct, since my final answer was 1 + 9b^2 . However, the actual answer was 1 + 6b + 9b^2 and I could not figure out why. While I was going over the homework with my friends and discussing this one question, the solution suddenly popped into my head and the question finally made sense!

I had been adding and multiplying the exponents together, but that was only half the battle:

(1^2 + 3^2b^2) = 1 + 3b^2

But the exponent meant that the binomial is multiplying by itself, so using distribution to solve the problem is the correct way to do it:

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(1 + 3b)\cdot (1 + 3b) = 1 + 6b + 9b^2

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Week 7 – Math 10

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This question stumped me for a very long time, since it had multiple parts that weren’t all trigonometry. For example, I was not sure if I had to use the distance over time triangle from physics, or if I should use unit analysis from the measurement unit. I tried many different things, and I finally found out how to solve the problem!

At first I used sine to figure out the length of the hypotenuse, but later on as I thought about it, I realized it would not make sense. The biker is going along a highway, it’s impossible to travel along the hypotenuse side! So I changed the equations:

tan 8 = \frac{60}{x}

\frac{60}{tan 8} = 426.92

\frac{426.92 km}{20 s}={x km}{1 hr}

I multiplied 20s by 3 into 60s, which is 1 minute. Then I multiplied it by 60 to convert it into 1 hour. You must do the same to the top numbers, so I got an answer of:

426.92\cdot 30\cdot 60 = 76.84

So the average speed the biker travelled at is 77km/hr!

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Solving Trig Equations

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sin x = \frac{5}{13}

For this triangle, the degree is the variable. To find the value of x, first we will need to label the sides:

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Sine is being used to solve the equation because the measurements of the hypotenuse and opposite sides are provided.

To isolate x, use algebra to cancel out the sine on the left side. What you do to one side, you must do to the other, so the equation becomes:

x = sin^{-1}\cdot \frac{5}{13}

Plug those numbers in on your calculator and you get x = 23 , which is the answer!

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tan 39 = \frac{17}{x}

This time, a degree and only one side length is given. In this case, x is the denominator of the fraction, which makes the problem a bit more complicated.

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We know that when we switch two fractions around, like so:

\frac{1}{2} = \frac{3}{6} —> \frac{6}{2} = \frac{3}{1}

They are equivalent. So, we can use this trick to make the equation easier to solve:

\frac{tan 39}{1} = \frac{17}{x} —> \frac{x}{1} = \frac{17}{tan 39}

x = 21

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cos 22 = \frac{x}{50}

The x is the numerator in this equation. So you will have to solve this equation  in a different way compared to the last two.

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Since the x is being divided by 50, if we multiply it by 50 the 50 will be canceled out. This means we also have to multiply the other side by 50:

50\cdot cos 22 = x

x = 46