Factorising quartics One solution of the quartic equation z4 + 2z³ + 2z² + 10z + 25 = 0 is z = -2 + i. Solve the equation.

Factorising quartics Since z = -2 + i is a solution, another solution is the complex conjugate z = -2 - i . Therefore two factors of the quartic expression are (z + 2 – i) and (z + 2 + i). So a quadratic factor is (z + 2 – i)(z + 2 + i). Multiplying out gives (z + 2)² - (i)² = z² + 4x + 4 - (-1) = z² + 4x + 5

Factorising quartics Now you need to factorise the quartic expression z4 + 2z³ + 2z² + 10z + 25 into two quadratic factors, where one factor is z² + 4z + 5.

Factorising polynomials This PowerPoint presentation demonstrates three methods of factorising a quartic into two quadratic factors when you know one quadratic factor. Click here to see factorising by inspection Click here to see factorising using a table Click here to see polynomial division

Factorising by inspection Write the unknown quadratic as az² + bz + c. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c)

Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(az² + bz + c) So a must be 1.

Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in z4 is by multiplying z2 by az2, giving az4. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(1z² + bz + c) So a must be 1.

Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + c) So c must be 5.

Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying 5 by c, giving 5c. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5) So c must be 5.

Factorising by inspection Now think about the term in z. When you multiply out the brackets, you get two terms in z. 4z multiplied by 5 gives 20z z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² + bz + 5) 5 multiplied by bz gives 5bz So 20z + 5bz = 10z therefore b must be -2.

Factorising by inspection Now think about the term in z. When you multiply out the brackets, you get two terms in z. 4z multiplied by 5 gives 20z z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5) 5 multiplied by bz gives 5bz So 20z + 5bz = 10z therefore b must be -2.

Factorising by inspection You can check by looking at the z² term. When you multiply out the brackets, you get three terms in z². z² multiplied by 5 gives 5z² z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5) 4z multiplied by -2z gives -8z² 5 multiplied by z² gives 5z² 5z² - 8z² + 5z² = 2z² as it should be!

Factorising by inspection Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5) The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.

Factorising polynomials Click here to see this example of factorising by inspection again Click here to see factorising using a table Click here to see polynomial division 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 Write the unknown quadratic as az² + bz + c. z² 4z 5 az² bz c

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 az² bz c z4 The only z4 term appears here, so this must be z4.

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 az² bz c z4 This means that a must be 1.

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 1z² bz c z4 This means that a must be 1.

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz c z4 25 The constant term, 25, must appear here

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz c z4 25 so c must be 5

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz 5 z4 25 so c must be 5

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz 5 z4 5z² 4z³ 20z 5z² 25 Four 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 z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz 5 z4 -2z³ 5z² 4z³ 20z 5z² 25 This space must contain an z³ term and to make a total of 2z³, this must be -2z³

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² bz 5 z4 -2z³ 5z² 4z³ 20z 5z² 25 This shows that b must be -2

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² -2z 5 z4 -2z³ 5z² 4z³ 20z 5z² 25 This shows that b must be -2

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² -2z 5 z4 -2z³ 5z² 4z³ -8z² 20z 5z² -10z 25 Now the last spaces in the table can be filled in

Factorising using a table The result of multiplying out using this table has to be z4 + 2z³ + 2z² + 10z + 25 z² 4z 5 z² -2z 5 z4 -2z³ 5z² 4z³ -8z² 20z 5z² -10z 25 and you can see that the term in z²is 2z² and the term in z is 10z, as they should be.

Factorising by inspection Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5) The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.

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

Algebraic long division Divide z4 + 2z³+ 2z² + 10z + 25 by z² + 4z + 5 z² + 4z + 5 is the divisor z4 + 2z³ + 2z² + 10z + 25 is the dividend The quotient will be here.

Algebraic long division First divide the first term of the dividend, z4, by z² (the first term of the divisor). z² This gives z². This will be the first term of the quotient.

Algebraic long division z² z4 + 4z³ + 5z² Now multiply z² by z² + 4z + 5 -2z³ - 3z² and subtract

Algebraic long division z² + 10z z4 + 4z³ + 5z² -2z³ - 3z² Bring down the next term, 10z

Algebraic long division z² - 2z z4 + 4z³ + 5z² Now divide -2z³, the first term of -2z³ - 3z² + 5, by z², the first term of the divisor -2z³ - 3z² + 10z which gives -2z

Algebraic long division z² - 2z z4 + 4z³ + 5z² -2z³ - 3z² + 10z Multiply -2z by z² + 4z + 5 -2z³- 8z²- 10z 5z²+ 20z and subtract

Algebraic long division z² - 2z + 25 z4 + 4z³ + 5z² -2z³ - 3z² + 10z Bring down the next term, 25 -2z³- 8z²- 10z 5z²+ 20z

Algebraic long division z² - 2z + 5 z4 + 4z³ + 5z² Divide 5z², the first term of 5z² + 20z + 25, by z², the first term of the divisor -2z³ - 3z² + 10z -2z³- 8z²- 10z 5z²+ 20z + 25 which gives 5

Algebraic long division z² - 2z + 5 z4 + 4z³ + 5z² -2z³ - 3z² + 10z Multiply z² + 4z + 5 by 5 -2z³- 8z²- 10z 5z²+ 20z + 25 Subtracting gives 0 as there is no remainder. 5z² + 20z + 25

Factorising by inspection Now you can solve the equation by applying the quadratic formula to z²- 2z + 5 = 0. z4 + 2z³ + 2z² +10z + 25 = (z² + 4z + 5)(z² - 2z + 5) The solutions of the equation are z = -2 + i, -2 - i, 1 + 2i, 1 – 2i.

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