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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Chapter 1 Algebra and functions

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Chapter 1 algebra and functions

Chapter 1 Algebra and functions

C2


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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