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Factorising polynomials
This PowerPoint presentation demonstrates three methods of factorising a polynomial when you know one linear factor.Click here to see factorising by inspection
Click here to see factorising using a table
Click here to see polynomial division
If you divide x³ - x² - 4x – 6 (cubic) by x – 3 (linear), then the result must be quadratic.
Write the quadratic as ax² + bx + c.
x³ – x² – 4x - 6 = (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 1.
Factorising by inspection
x³ – x² – 4x - 6 = (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 1.
Factorising by inspection
x³ – x² – 4x - 6 = (x – 3)(1x² + 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 2.
Factorising by inspection
x³ – x² – 4x - 6 = (x – 3)(x² + 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 2.
Factorising by inspection
x³ – x² – 4x - 6 = (x – 3)(x² + bx + 2)
Now think about the x² term. When you multiply out the brackets, you get two x² terms.
-3 multiplied by x² gives –3x²
x multiplied by bx gives bx²
So –3x² + bx² = -1x²therefore b must be 2.
x³ – x² – 4x - 6 = (x – 3)(x² + bx + 2)
Factorising by inspection
Now think about the x² term. When you multiply out the brackets, you get two x² terms.
-3 multiplied by x² gives –3x²
x multiplied by bx gives bx²
So –3x² + bx² = -1x²therefore b must be 2.
x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)
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 2x gives -6x
x multiplied by 2 gives 2x
-6x + 2x = -4x as it should be!
x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)
Factorising by inspection
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to x²+ 2x + 2 = 0.
x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2)
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.
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
Click here to see polynomial division
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 x³ - x² - 4x - 6 (cubic) by x – 3 (linear), then the result must be quadratic.Write the quadratic as ax² + bx + c.
Factorising using a table
x
-3
ax² bx c
x
-3
ax² bx c
The result of multiplying out using this table has to be x³ - x² - 4x - 6
The only x³ term appears here,
so this must be x³.
x³
Factorising using a table
x
-3
ax² bx c
The result of multiplying out using this table has to be x³ - x² - 4x - 6
This means that a must be 1.
x³
Factorising using a table
x
-3
1x² bx c
The result of multiplying out using this table has to be x³ - x² - 4x - 6
This means that a must be 1.
x³
Factorising using a table
x
-3
x² bx c
The result of multiplying out using this table has to be x³ - x² - 4x - 6
The constant term, -6, must appear here
x³
Factorising using a table
-6
x
-3
x² bx c
The result of multiplying out using this table has to be x³ - x² - 4x - 6
so c must be 2
x³
Factorising using a table
-6
x
-3
x² bx 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
so c must be 2
x³
Factorising using a table
-6
x
-3
x² bx 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
Two more spaces in the table can now be filled in
x³
Factorising using a table
-6
2x
-3x²
x
-3
x² bx 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
This space must contain an x² term
x³
Factorising using a table
-6
2x
-3x²
and to make a total of –x², this must be 2x²
2x²
x
-3
x² bx 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
This shows that b must be 2
x³
Factorising using a table
-6
2x
-3x²
2x²
x
-3
x² 2x 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
This shows that b must be 2
x³
Factorising using a table
-6
2x
-3x²
2x²
x
-3
x² 2x 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
Now the last space in the table can be filled in
x³
Factorising using a table
-6
2x
-3x²
2x²
-6x
x
-3
x² 2x 2
The result of multiplying out using this table has to be x³ - x² - 4x - 6
and you can see that the term in x is -4x, as it should be.
x³
Factorising using a table
-6
2x
-3x²
2x²
-6x
So x³ - x² - 4x - 6 = (x – 3)(x² + 2x + 2)
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0.
x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2)
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.
Factorising polynomials
Click here to see factorising by inspection
Click here to see this example of factorising using a table again
Click here to end the presentation
Click here to see polynomial division
Algebraic long division
Divide x³ - x² - 4x - 6 by x - 3
3 23 4 6x x x xx - 3 is the divisor
The quotient will be here.
x³ - x² - 4x - 6 is the dividend
Algebraic long division
3 23 4 6x x x x
First divide the first term of the dividend, x³, by x (the first term of the divisor).
This gives x². This will be the first term of the quotient.
x²
Algebraic long division
3 23x x
3 23 4 6x x x xx²
22x - 4x
Now divide 2x², the first term of 2x² - 4x, by x, the first term of the divisor
which gives 2x
+ 2x
Algebraic long division
3 23x x
3 23 4 6x x x xx²
22x - 4x
+ 2x
Multiply 2x by x - 3
and subtract
2x² - 6x
2x
Algebraic long division
3 23x x
3 23 4 6x x x xx²
22x - 4x
+ 2x
2x² - 6x
2x
Bring down the next term, -6
- 6
Algebraic long division
3 23x x
3 23 4 6x x x xx²
22x - 4x
+ 2x
2x² - 6x
2x - 6
Divide 2x, the first term of 2x - 6, by x, the first term of the divisor which gives 2
+ 2
Algebraic long division
3 23x x
3 23 4 6x x x xx²
22x - 4x
+ 2x
2x² - 6x
2x - 6
+ 2
Multiply x - 3 by 2
Subtracting gives 0 as there is no remainder.
2x - 60
Factorising by inspection
Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0.
So x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2)
The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.