Factoring and Radicals Presentation !. By: Amandeep , Varun , and Vinit Due Date: November 11th 2011 Class Course: MPM1D8-D Teacher : Anderson, J. Presentation Content. Simplifying Radicals (Surds) Expanding and Simplifying Radicals( Surds ) Simple Trinomials Common Factors
By: Amandeep, Varun, and Vinit
Due Date: November11th 2011
Class Course: MPM1D8-D
Teacher: Anderson, J
2 x √ 3 x 4 x √3
You multiply under square root by under square root and number with number. This gives you:
8 x √9
=8 x 3
What if you have something like
What you do is you find largest square factor in the number.
You then find its square root.
As you can see, we leave the number under square root that cannot be simplified without giving an irrational number.
(3 + √2)(3 + √8)
When you multiply through, you multiply under square root with under square root (remember √n has a one in front of it. It just doesn’t show) and number with number. You then get:
9 + 3√8 + 3√2 + √16 simplify
= 9 + 3√8 + 3√2 + 4 collect like terms so in this case number and number. Leave the under square root because they are not like terms unless the number under square root are the same
= 14 +3√8 + 3√2
To factor a number or an expression, means to write it as multiplication, that is, as a product of factors.
Factoring is the reverse of multiplying. When we multiply, we write
2(a + b) = 2a + 2b.
But if we switch sides and write
2a + 2b = 2(a + b),
then we have factored
2a + 2b as the product
2(a + b).
you must first find the highest common factor.
The highest common factor (HCF) will contain the lowest power of each letter.Practice a) 3abc − 4ab = b) 2xy − 8xyz = c) x²y3 −x3y² = d) 8ab3+12a²b²=
For example :
x²y3z4 + x4yz3=
The HCF is x²y z3
because in the equation we must find the lowest power for each letter, this will give you what to factor from equation.
It is always good to double check your work
If we multiply the right-hand side, we will obtain the left-hand side.
Group the first and second terms - find their common factor. Do the same with the third and fourth terms.
x3 − 5x² + 3x − 15
= x²(x − 5) + 3(x − 5)
= (x² + 3)(x − 5).
a) x3 + x² + 3x + 3 =
b) 2x3 − 6x² + 5x − 15 =
c) 3x3 − 15x² − 2x + 10 =
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Remember from your translation skills that "difference" means "subtraction". So a difference of squares is something that looks like x2 – 4. That's because 4 = 22, so you really have x2 – 22, a difference of squares. To factor this, do your parentheses, same as usual:
x2 – 4 = (x )(x )
You need factors of –4 that add up to zero, so use –2 and +2:
x2 – 4 = (x – 2)(x + 2)
Note that we had x2 – 22, and ended up with (x – 2)(x + 2). Differences of squares (something squared minus something else squared) always work this way:
For a2 – b2, do the parentheses:
( )( )
...put the first squared thing in front:
(a )(a )
...put the second squared thing in back:
...and alternate the signs in the middles:
(a – b)(a + b)
Memorize this formula! It will come in handy
Factor x4 – 1
This is (x2)2 – 12, so I get:
x4 – 1 = (x2)2 – 12 = (x2 – 1)(x2 + 1)
Note that I'm not done yet, because x2 – 1 is itself a difference of squares, so I need to apply the formula again to get the fully-factored form. Since x2 – 1 = (x – 1)(x + 1), then:
x4 – 1 = (x2)2 – 12 = (x2 – 1)(x2 + 1)
= ((x)2 – (1)2)(x2 + 1)
= (x – 1)(x + 1)(x2 + 1)
A "quadratic" is a polynomial that looks like "ax2 + bx + c", where "a", "b", and "c" are just numbers. For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term "c", but also add up to equal "b", the coefficient on the x-term. For instance:
Factor x2 + 5x + 6.
I need to find factors of 6 that add up to 5. Since 6 can be written as the product of 2 and 3, and since 2 + 3 = 5, then I'll use 2 and 3. I know from multiplying polynomials that this quadratic is formed from multiplying two factors of the form "(x + m)(x + n)", for some numbers m and n. So I'll draw my parentheses, with an "x" in the front of each:
Then I'll write in the two numbers that I found before:
(x + 2)(x + 3)
This is the answer: x2 + 5x + 6 = (x + 2)(x + 3)
Note that you can always check your work by multiplying back to get the original answer.
In this case:
If cis positive, then the factors you're looking for are either both positive or else both negative.If bis positive, then the factors are positiveIf b is negative, then the factors are negative.If c is negative, then the factors you're looking for are of alternating signs;that is, one is negative and one is positive.If b is positive, then the larger factor is positive.If b is negative, then the larger factor is negative.