Remainder and Factor Theorem Name : __________________( ) Class : _________ Date : _________

3x3 4x2 x 2 1 1 1 1 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. 3x3 4x2 x 2 1 1 1 1

Subtract (which changes the sign of each term in the polynomial) 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 and copy the steps. Subtract (which changes the sign of each term in the polynomial) Now multiply by the divisor and put the answer below. x + 11 2 1 Bring down the next number or term Multiply and put below Now divide 3 into 5 or x into 11x First divide 3 into 6 or x into x2 32 698 x - 3 x2 + 8x - 5 Remainder added here over divisor 64 x2 – 3x subtract 5 8 11x - 5 32 11x - 33 This is the remainder 26 28 So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way:

Remainder added here over divisor 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 changes the sign of each term in the polynomial) y - 2 Write out with long division including 0y for missing term Multiply and put below Bring down the next term Multiply and put below y + 2 y2 + 0y + 8 Divide y into y2 Divide y into -2y Remainder added here over divisor y2 + 2y subtract -2y + 8 - 2y - 4 This is the remainder 12

Reminder: For a polynomial function , the factor theorem says that: if then is a factor e.g. 1 Find one linear factor of is a factor

x and 2x are linear factors of Now suppose Possible linear factors are now x and 2x are linear factors of

x and 2x are linear factors of Now suppose Possible linear factors are now x and 2x are linear factors of 1 and 3 are the factors of 3

x and 2x are linear factors of Now suppose Possible linear factors are now x and 2x 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,

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

So, We can now complete the factorisation by inspection:

So, We can now complete the factorisation by inspection:

So, We can now complete the factorisation by inspection:

So, We can now complete the factorisation by inspection: Checking the quadratic term:

So, We can now complete the factorisation by inspection: Checking the quadratic term:

So, We can now complete the factorisation by inspection: Checking the linear term:

So, We can now complete the factorisation by inspection: Checking the linear term:

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

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.

We can start with . So,

We can start with . So,

We can start with . So,

We can start with . So,

We can start with . So, Check the linear term

We can start with . So, Check the linear term

We can start with . So, There are no more factors.

SUMMARY 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.

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

1. Let Possible factors are Solution: So,

So, There are no more factors. 2. Let Solution: Show is a factor So,

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 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 is divided by ( x - 1 ). 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

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

SUMMARY Extending the Remainder theorem The remainder when a polynomial is divided by is given by R where

We can start with . So, There are no more factors.

To find the remainder when a polynomial is divided by solve to find the value of x let substitute into to find the remainder

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