The Factor and Remainder Theorems Revisited Remainder and

The Factor and Remainder Theorems Revisited Remainder and Factor Theorem Name : _________( Class : _____ Date : _____ )

The Factor and Remainder Theorems Revisited Dividing by a Monomial If the divisor only has one term, split the polynomial up into a fraction for each term. divisor Now reduce each fraction. 3 x 3 4 x 2 1 1

The Factor and Remainder Theorems Revisited Long Division If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials Subtract (which and copy 2 the Now multiply by x + 11 1 steps. changes the sign Bring down the divisor and put Multiply and 2 + 8 x - 5 32 698 x 3 x of each term in next number or the answer below. put below Remainder 2 – 3 x 64 x the polynomial) term subtract added here 58 3 into 6 First divide or x into x 2 11 x - 5 over divisor 32 Now divide is the x into 11 x - 33 26 3 into 5 This or remainder 28 So we found the answer to the problem x 2 + 8 x – 5 x – 3 or the problem written another way:

The Factor and Remainder Theorems Revisited Let's Try Another One If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight. Subtract (which y-2 Write out with changes theand sign Multiply Bring Multiply down and the 2 + 0 y + 8 long division 2 y + 2 y Divide intoiny-2 y Divide yyinto of each term put below next put term below including 0 y for Remainder 2 + 2 y y the polynomial) subtract missing term added here This is the remainder -2 y + 8 over divisor - 2 y - 4 12

The Factor and Remainder Theorems Revisited Reminder: For a polynomial function says that: if , the factor theorem then is a factor e. g. 1 Find one linear factor of is a factor

The Factor and Remainder Theorems Revisited Now suppose Possible linear factors are now x and 2 x are linear factors of

The Factor and Remainder Theorems Revisited Now suppose Possible linear factors are now x and 2 x are linear factors of 1 and 3 are the factors of 3

The Factor and Remainder Theorems Revisited Now suppose Possible linear factors are now x and 2 x are linear factors of 1 and 3 are the factors of 3 To check if is a factor we use x = 1 The value of x comes from letting So, to check we let So,

The Factor and Remainder Theorems Revisited So, to factorise Consider: So, Starting with the easiest: Can you see that either? ( The only difference is that the signs of the 1 st and 3 rd terms change. )

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection: Checking the quadratic term:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection: Checking the quadratic term:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection: Checking the linear term:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection: Checking the linear term:

The Factor and Remainder Theorems Revisited So, We can now complete the factorisation by inspection: The quadratic factor has no linear factors, so the factorisation is complete.

The Factor and Remainder Theorems Revisited e. g. 2 Factorise Solution: Let Possible factors are So, BUT we can save time by noticing that, since all the terms of are positive, and cannot be factors. Also, you may have spotted that the coefficients show that isn’t a factor.

The Factor and Remainder Theorems Revisited We can start with So, .

The Factor and Remainder Theorems Revisited We can start with So, .

The Factor and Remainder Theorems Revisited We can start with So, .

The Factor and Remainder Theorems Revisited We can start with So, .

The Factor and Remainder Theorems Revisited We can start with So, Check the linear term .

The Factor and Remainder Theorems Revisited We can start with So, Check the linear term .

The Factor and Remainder Theorems Revisited We can start with . So, There are no more factors.

The Factor and Remainder Theorems Revisited SUMMARY Ø Factorising Cubic Functions • Use the factor theorem to find one linear factor For checking for a factor of the form let and solve to find the value of x to substitute. • Use inspection to find the quadratic factor. • Check for further factors of the quadratic.

The Factor and Remainder Theorems Revisited Exercises 1. Factorise the following cubic: 2. Show that is a factor of the following cubic and then factorise it completely:

The Factor and Remainder Theorems Revisited 1. Let Solution: Possible factors are So,

The Factor and Remainder Theorems Revisited So, There are no more factors. 2. Let Solution: Show So, is a factor

The Factor and Remainder Theorems Revisited The slides that follow remind you about the remainder theorem. The theorem is extended to consider linear terms of the form in the same way as the factor theorem.

The Factor and Remainder Theorems Revisited The remainder theorem gives the remainder when a polynomial is divided by a linear factor It doesn’t enable us to find the quotient. e. g. 1 Find the remainder when divided by ( x - 1 ). is The method is the same as that for the factor theorem: The remainder is 4 The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by

The Factor and Remainder Theorems Revisited e. g. 2 Find the remainder when is divided by. Solution: Let To find the value of a, we let 2 x + 1 = 0. The value of x gives the value of a. so,

The Factor and Remainder Theorems Revisited SUMMARY Ø Extending the Remainder theorem • The remainder when a polynomial by is given by R where is divided

The Factor and Remainder Theorems Revisited We can start with . So, There are no more factors.

The Factor and Remainder Theorems Revisited To find the remainder when a polynomial divided by • let • solve to find the value of x • substitute into is to find the remainder

The Factor and Remainder Theorems Revisited e. g. Find the remainder when divided by. is Solution: Let To find the value of a, we let 2 x + 1 = 0. The value of x gives the value of a. so,