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 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. - PowerPoint PPT Presentation
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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
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.
So a must be 2.
Factorising by inspection
2x³ – 5x² – 4x + 3 = (x – 3)(ax² + bx + c)
Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³.
So a must be 2.
Factorising by inspection
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx + c)
Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c.
So c must be -1.
Factorising by inspection
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx + c)
Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c.
So c must be -1.
Factorising by inspection
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 1)
Now think about the x² term. When you multiply out the brackets, you get two x² terms.
-3 multiplied by 2x² gives –6x²
x multiplied by bx gives bx²
So –6x² + bx² = -5x²therefore b must be 1.
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + bx - 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
x multiplied by bx gives bx²
So –6x² + bx² = -5x²therefore b must be 1.
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + 1x - 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
x multiplied by –1 gives -x
-3x – x = -4x as it should be!
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)
Factorising by inspection
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)
Factorising by inspection
Now factorise the quadratic in the usual way.
= (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
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 - 12The 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
x
-3
ax² bx c
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
The only x³ term appears here,
so this must be 2x³.
2x³
Factorising using a table
This means that a must be 2.
x
-3
ax² bx c
2x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
This means that a must be 2.
x
-3
2x² bx c
2x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
The constant term, 3, must appear here
3
x
-3
2x² bx c
2x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
so c must be –1.
3
x
-3
2x² bx c
2x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
so c must be –1.
3
x
-3
2x² bx -12x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
3
x
-3
2x² bx -12x³
Two more spaces in the table can now be filled in
-6x²
-x
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
This space must contain an x² term
and to make a total of –5x², this must be x²
x²
3
x
-3
2x² bx -12x³
-6x²
-x
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
This shows that b must be 1.
x²
3
x
-3
2x² bx -12x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
-6x²
-x
This shows that b must be 1.
x²
3
x
-3
2x² 1x -12x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
Factorising using a table
-6x²
-x
Now the last space in the table can be filled in
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
x²
-3x 3
x
-3
2x² x -1
2x³
-6x²
-x
Factorising using a table
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)
x²
3
x
-3
2x² x -1
2x³
The result of multiplying out using this table has to be 2x³ - 5x² - 4x + 3
-6x²
-x
Factorising using a table
-3x
2x³ – 5x² – 4x + 3 = (x – 3)(2x² + x - 1)
Factorising by inspection
Now factorise the quadratic in the usual way.
= (x – 3)(2x – 1)(x + 1)
Factorising polynomials
Click here to see this example of factorising using a table again