chapter 1 algebra and functions
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Chapter 1 Algebra and functions. C2. Example 1. Simplify this expression 4x 4 +5x 2 -7x x Here write these as three separate fractions. Another example. Simplify the following n 2 +8n +16 n 2 –16

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example 1
Example 1
  • Simplify this expression
  • 4x4+5x2-7x
  • x
  • Here write these as three separate fractions
another example
Another example
  • Simplify the following n2+8n +16
  • n2 –16
  • This is a perfect square and a difference of two squares- cookie cutters!
  • = (n+4)2
  • (n-4)(n+4)
  • = (n+4)
  • (n-4)
dividing polynomials by cancelling
Dividing polynomials by cancelling
  • Always factorise first using the HCF and then cancel out common factors in the fraction
  • Example (2c2d3)2
  • 8b4c4
  • Quiz
  • Some classwork questions
the family division
The family division
  • Remember your family
  • Dad
  • Mum
  • Sister
  • Brother
example 1 no remainder
Example 1 No remainder
  • Set you work out using the family steps
  • Divide x3 + 2 x2 -17x +6 by (x-3)
  • x-3 ) x3+2x2–17x +6
no remainder again
No remainder again
  • Divide 6x3+28x2-7x+15 by (x+5)
leaving a gap
Leaving a gap!
  • Divide x3-3x-2 by (x-2)
example with a remainder
Example with a remainder
  • Divide 2x3–5x2-16x +10 by (x-4)
  • Now we must write the polynomial with highest power first!
  • x –4 ) 2x3 -5x2 –16x+ 10
a summary of long division
A summary of long division
  • Always follow Dad, mum, sister, brother
  • Always write the polynomial in order starting with the highest power
  • Always leave a space for any terms (powers) not in the question
  • More worked examples?
  • Click on the picture!
factor theorem
Factor theorem
  • if x-a is a factor of f(x) then f(a)=0
  • What does this mean?
  • If (x-2) is a factor of
  • f(x) = x3+x2-4x-4 then f(2) = 0. please check this.
  • F(2) = 8 +4-8-4
factor theorem example 2
Factor Theorem example 2
  • Show the (x-1) is a factor of x3+6x2+5x-12 and hence fully factorize the expression.
example 3
Example 3
  • Given that (x+1) is a factor of

4x4-3x2+a find the value of a.

  • F(-1) = 4-3+a
  • 1 + a = 0
  • a = -1
example 4
Example 4
  • Prove that (2x+1) is a factor of 2x3+x2-18x-9
  • What do you substitute here?
  • Here we substitute x = -1/2
  • F(-1/2) = 2 (-1/8) + ¼ +9-9
  • = 0
finding factors of a polynomial
Finding factors of a polynomial
  • Fully factorise f(x) = 2x3+x2-18x-9
  • The first step here is to look at the constant 9 and try the factors of 9.
  • We will try f(1), f(-1), f( 3) and see which one equals 0.
  • F(1) = 2+1-18-9
  • F(3) = 54 + 9 -54 – 9!
  • So (x-3) is a factor.
  • Now we use family division to find the other factors.
  • Ex 1D.
classwork
Classwork
  • Just one more practice question on Sos maths. Let’s do this together.
remainder theorem
Remainder Theorem
  • If when you substitute f(a) into the polynomial and it does not equal zero then this number is actually the remainder.
  • Example find the remainder when x3-20x+3 is divided by (x-4)
  • F(4) = -13
  • Ex 1E Mixed exercise 1F
a competition
A competition
  • 1) Show that x-2 is a factor of
  • f(x) = x3+x2-5x-2 and hence or otherwise find the exact solutions of the equation f(x) = 0
a competition1
A competition
  • 2) Given that x = -1 is a root of the equation 2x3-5x2-4x+3, find the other two positive roots.
a competition2
A competition
  • 3) H(x) = x3+4x2+rx+s. Given that
  • H(-1) = 0 and H(2) = 30, find the values of r and s. Find the remainder when H(x) is divided by (3x-1)
a competition3
A competition
  • 4) Given that g(x) = 2x3+9x2-6x-5, factorise g(x) and solve g(x) = 0 .
a competition4
A competition
  • 5) F(x) = 2x2+px+q. Given that

F(-3)=0 and F(4) = 2 find the value of p and q and hence fully factorise.

five quick questions
Five quick questions
  • Write five quick quiz questions on the sheet provided
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