Chapter 1 Algebra and functions

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# Chapter 1 Algebra and functions - PowerPoint PPT Presentation

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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### Chapter 1 Algebra and functions

C2

Example 1
• Simplify this expression
• 4x4+5x2-7x
• x
• Here write these as three separate fractions
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
• 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
• Mum
• Sister
• Brother
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
• Divide 6x3+28x2-7x+15 by (x+5)
Leaving a gap!
• Divide x3-3x-2 by (x-2)
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
• 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
• 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
• Show the (x-1) is a factor of x3+6x2+5x-12 and hence fully factorize the expression.
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
• 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
• 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
• Just one more practice question on Sos maths. Let’s do this together.
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
• 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 competition
• 2) Given that x = -1 is a root of the equation 2x3-5x2-4x+3, find the other two positive roots.
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 competition
• 4) Given that g(x) = 2x3+9x2-6x-5, factorise g(x) and solve g(x) = 0 .
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
• Write five quick quiz questions on the sheet provided