quad equations

i have a couple of expressions i've had trouble with:
i'm trying to factorise

2x^4 + 14x^2 + 24

6x^2 -10x + 4

would be grateful if someone could explain and show working to this.
i'm sure pretty straight forward for some of you...

Reply 1

TenOfThem

Original post by alevelstud

i have a couple of expressions i've had trouble with:
i'm trying to factorise

2x^4 + 14x^2 + 24

Take out a factor of 2

Consider wether you can factorise

x^2 + 7x + 12

If you can then look for the connection between my equation and yours

Reply 2

tombayes

Original post by alevelstud

For the first one let

z=x^2

then it becomes

2z^2+14z+24

Reply 3

TenOfThem

Original post by tombayes

For the first one let

z=x^2

then it becomes

2z^2+14z+24

There is no need to make a substitution

IMO that simply complicates the question

I see too many weaker students who "forget" that they have substituted and then answer the wrong question

Whilst there is a need to think about the quadratic nature of the expression I think that introducing z is the wrong way to go

NB I know that the substitution method is favoured by many on here - just not by me

Reply 4

SteveMcs

Original post by TenOfThem

Take out a factor of 2

Consider wether you can factorise

x^2 + 7x + 12

If you can then look for the connection between my equation and yours

Surely a factor of 2 can only be taken out if it's equal to zero?

either way it's not necessary in the case because as it is, it factorises nicely to (2x^2 + 8)(x^2 + 3)

Reply 5

TenOfThem

Original post by SteveMcs

Surely a factor of 2 can only be taken out if it's equal to zero?

either way it's not necessary in the case because as it is, it factorises nicely to @@@@

Seriously?

you think that (for example) 2x+4 = 2(x+2) is not a possible factorisation because there was no equation = 0?

And, please do not give "full" solutions - even when they are incorrect

Reply 6

tombayes

Original post by SteveMcs

Surely a factor of 2 can only be taken out if it's equal to zero?

You cannot divide by two but you can take it outside the brackets

e.g. 2x^2+2 can become 2(x^2+1)

Reply 7

SteveMcs

Original post by TenOfThem

Seriously?

you think that (for example) 2x+4 = 2(x+2) is not a possible factorisation because there was no equation = 0?

And, please do not give "full" solutions - even when they are incorrect

I misunderstood when you said take a factor of 2 out, i thought you meant cancel the 2 out (which you would only be able to do if the equation was set to 0) so i apologize.

But i don't think my solution was incorrect.. When my factorization is multiplied out it gives the equation i started with (Obviously if it is incorrect please tell the OP how so that they don't make the same mistake)

Reply 8

TenOfThem

Original post by SteveMcs

I misunderstood when you said take a factor of 2 out, i thought you meant cancel the 2 out (which you would only be able to do if the equation was set to 0) so i apologise.

Fair enough

But i don't think my solution was incorrect.. When my factorization is multiplied out it gives the equation i started with (Obviously if it is incorrect please tell the OP how so that they don't make the same mistake)

Would you consider 4x+8 = 2(2x+4) to be the correct answer if asked to factorise - when asked to factorise there is an implied "fully"

Reply 9

alevelstud

thanks for the replies and good to see some debate going on.
any ideas on the second one

6x^2 -10x + 4

Reply 10

davros

Original post by alevelstud

thanks for the replies and good to see some debate going on.
any ideas on the second one

6x^2 -10x + 4

Again, take out a factor of 2 first to make things a bit more manageable, then look at factorizing what you've got left.

What techniques do you know for factorizing quadratics when the coefficient of x^2 isn't 1?

Reply 11

alevelstud

Original post by davros

thanks.

i've only started doing maths again after more than 6 years. i can solve equations of the form ax^2 + bx +c using 2 factors of ac which add up to make b.