Factorising polynomials This PowerPoint 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 2x³ - 5x² - 4x – 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. 2x³ – 5x² – 4x + 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. 2x³ – 5x² – 4x + 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. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 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 –3c. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 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 –3c. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 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 2x² gives –6x² 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 1) x multiplied by bx gives bx² So –6x² + bx² = -5x² therefore b 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 2x² gives –6x² 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 1x - 1) x multiplied by bx gives bx² So –6x² + bx² = -5x² therefore b must be 1.

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 –3x 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1) x multiplied by –1 gives -x -3x – x = -4x as it should be!

Factorising by inspection Now factorise the quadratic in the usual way. 2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1) = (x – 3)(2x – 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: 2x 3 x² -3x - 4 2x³ -6x² -8x 3x² -9x -12 So (2x + 3)(x² - 3x – 4) = 2x³ - 3x² - 17x - 12 The method you are going to see now is basically the reverse of this process.

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

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

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

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

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

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

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

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

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

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

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

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