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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

Chapter 1 Algebra and functions

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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

• 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