Factorising polynomials This Power Point presentation demonstrates two

Factorising polynomials This Power. Point presentation demonstrates two methods of factorising a polynomial when you know one factor (perhaps by using the factor theorem). Click here to see factorising by inspection Click here to see factorising using a table

Factorising by inspection If you divide 2 x³ - 5 x² - 4 x – 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(ax² + bx + c)

Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(ax² + bx + c) So a must be 2.

Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + bx + c) So a must be 2.

Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying – 3 by c, giving – 3 c. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + bx + c) So c must be -1.

Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying – 3 by c, giving – 3 c. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + bx - 1) So c must be -1.

Factorising by inspection Now think about the x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by 2 x² gives – 6 x² 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + bx - 1) So – 6 x² + bx² = -5 x² therefore b must be 1. x multiplied by bx gives bx²

Factorising by inspection Now think about the x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by 2 x² gives – 6 x² 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + 1 x - 1) So – 6 x² + bx² = -5 x² therefore b must be 1. x multiplied by bx gives bx²

Factorising by inspection You can check by looking at the x term. When you multiply out the brackets, you get two terms in x. -3 multiplied by x gives – 3 x 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + x - 1) -3 x – x = -4 x as it should be! x multiplied by – 1 gives -x

Factorising by inspection Now factorise the quadratic in the usual way. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + x - 1) = (x – 3)(2 x – 1)(x + 1)

Factorising polynomials Click here to see this example of factorising by inspection again Click here to see factorising using a table Click here to end the presentation

Factorising using a table If you find factorising by inspection difficult, you may find this method easier. Some people like to multiply out brackets using a table, like this: x² 2 x 2 x³ 3 3 x² -3 x -4 -6 x² -9 x -8 x -12 So (2 x + 3)(x² - 3 x – 4) = 2 x³ - 3 x² - 17 x - 12 The method you are going to see now is basically the reverse of this process.

Factorising using a table If you divide 2 x³ - 5 x² - 4 x + 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. ax² x -3 bx c

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x ax² 2 x³ bx c -3 The only x³ term appears here, so this must be 2 x³.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x ax² 2 x³ bx c -3 This means that a must be 2.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x 2 x² 2 x³ bx c -3 This means that a must be 2.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x -3 2 x² 2 x³ bx c 3 The constant term, 3, must appear here

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x -3 2 x² 2 x³ bx c 3 so c must be – 1.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x -3 2 x² 2 x³ bx -1 3 so c must be – 1.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x 2 x² 2 x³ -3 -6 x² bx -1 -x 3 Two more spaces in the table can now be filled in

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x 2 x² 2 x³ -3 -6 x² bx x² -1 -x 3 This space must contain an x² term and to make a total of – 5 x², this must be x²

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x 2 x² 2 x³ -3 -6 x² bx x² -1 -x 3 This shows that b must be 1.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 x 2 x² 2 x³ -3 -6 x² 1 x x² -1 -x 3 This shows that b must be 1.

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 2 x² 2 x³ x x² -1 -x -3 -6 x² -3 x 3 x Now the last space in the table can be filled in

Factorising using a table The result of multiplying out using this table has to be 2 x³ - 5 x² - 4 x + 3 2 x² 2 x³ x x² -1 -x -3 -6 x² -3 x 3 x and you can see that the term in x is – 4 x, as it should be. So 2 x³ - 5 x² - 4 x + 3 = (x – 3)(2 x² + x – 1)

Factorising by inspection Now factorise the quadratic in the usual way. 2 x³ – 5 x² – 4 x + 3 = (x – 3)(2 x² + x - 1) = (x – 3)(2 x – 1)(x + 1)

Factorising polynomials Click here to see this example of factorising using a table again Click here to see factorising by inspection Click here to end the presentation