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AS maths c1 help watch

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    Hi guys, need help factorising this equation - also any tips on factorising equations that have a number in front of x^2 that would be very helpful and greatly appreciated! Thanks

    4x^2-8x+3=0
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    Multiply the coefficient of x^2 term by get constant term. In other words multiply a by c to get ac. Then find 2 numbers that multiply to make ac but add to make b - the coefficient of the x term. Then split the x term into whatever these 2 numbers are. There should now be 4 terms in the expression. Now factorise the first 2 terms and then factorise the last 2 terms separately. You should find that there are 2 identical brackets. Then take out the expression in the brackets as a common factor and you should have factorised the quadratic into 2 linear factors.
    I will give an example of this.
    If we have 2x^2 + 9x + 10
    We need to find 2 numbers that multiply to make 2*10 , 20 and add together to make 9 so we can see that the numbers 4 and 5 make this work.
    Now we split the x term - 9x into the 2 numbers that we just found.
    So 9x = 4x+5x
    So the quadratic becomes
    2x^2 + 4x+ 5x + 10
    We factorise 2x^2 + 4x and 5x+10 separately
    So we get 2x(x+2) +5(x+2)
    So you see we have 2 identical brackets (x+2) common to both the 2x term and the constant term 5
    Now if we take out (x+2) as a common factor we get (x+2)(2x+5)
    So now we see that
    2x^2 + 9x + 10 = (2x+5)(x+2)
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    (Original post by B_9710)
    Multiply the coefficient of x^2 term by get constant term. In other words multiply a by c to get ac. Then find 2 numbers that multiply to make ac but add to make b - the coefficient of the x term. Then split the x term into whatever these 2 numbers are. There should now be 4 terms in the expression. Now factorise the first 2 terms and then factorise the last 2 terms separately. You should find that there are 2 identical brackets. Then take out the expression in the brackets as a common factor and you should have factorised the quadratic into 2 linear factors.
    I will give an example of this.
    If we have 2x^2 + 9x + 10
    We need to find 2 numbers that multiply to make 2*10 , 20 and add together to make 9 so we can see that the numbers 4 and 5 make this work.
    Now we split the x term 9x into the 2 numbers that we just found.
    So 9x = 4x+5x
    So the quadratic becomes
    2x^2 + 4x+ 5x + 10
    We factorise 2x^2 + 4x and 5x+10 separately
    So we get 2x(x+2) +5(x+2)
    So you see we have 2 identical brackets (x+2) common to both the 2x term and the constant term 5
    Now if we take out (x+2) as a common factor we get (x+2)(2x+5)
    So now we see that
    2x^2 + 9x + 10 = (2x+5)(x+2)
    typos
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    (Original post by YsfAli)
    Hi guys, need help factorising this equation - also any tips on factorising equations that have a number in front of x^2 that would be very helpful and greatly appreciated! Thanks

    4x^2-8x+3=0
    The constant is prime so that makes it easier as we know the factors.
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    (Original post by )
    typos
    I was doing an example of a different quadratic so that I do not just give the answer as it is against forum rules.
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    (Original post by YsfAli)
    Hi guys, need help factorising this equation - also any tips on factorising equations that have a number in front of x^2 that would be very helpful and greatly appreciated! Thanks

    4x^2-8x+3=0
    Hi,
    I'll use a different example. 6x^2 - 7x - 3 = 0.

    First of all, find two numbers that will multiply to give 3. The only possible combination is 3 and 1. These go here:
    ( 3)( 1)= 0

    Next find two numbers that multiply to give 6. There are two possible combinations: 3 and 2 OR 6 and 1.

    Also, because the 3 has a minus in front of it, one of the brackets needs to have a plus and the other a minus, so that when you expand the brackets, you get the -3 term.
    You now try out the possible combinations until it works.

    The answer comes out to be (3x -1)(2x + 3)=0

    Note that sometimes you cannot factorise using this method and you need to use the quadratic formula or completing the square method instead.

    Your equation will work using the method I showed above.
    Hope this helps!
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    (Original post by B_9710)
    I was doing an example of a different quadratic so that I do not just give the answer as it is against forum rules.
    You said 4 and 5 multiply to give 20 and add to give 7, that's wrong; it's not a different quadratic its simply wrong.
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    (Original post by )
    You said 4 and 5 multiply to give 20 and add to give 7, that's wrong; it's not a different quadratic its simply wrong.
    Oh yeah. It's because I started off doing a different example but changed halfway through and forgot about the 7
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    (Original post by Angel28)
    Hi,
    I'll use a different example. 6x^2 - 7x - 3 = 0.

    First of all, find two numbers that will multiply to give 3. The only possible combination is 3 and 1. These go here:
    ( 3)( 1)= 0

    Next find two numbers that multiply to give 6. There are two possible combinations: 3 and 2 OR 6 and 1.

    Also, because the 3 has a minus in front of it, one of the brackets needs to have a plus and the other a minus, so that when you expand the brackets, you get the -3 term.
    You now try out the possible combinations until it works.

    The answer comes out to be (3x -1)(2x + 3)=0

    Note that sometimes you cannot factorise using this method and you need to use the quadratic formula or completing the square method instead.

    Your equation will work using the method I showed above.
    Hope this helps!
    That actually makes more sense to me as my teacher doesn't really bother explaining little things in detail. Thanks a lot really do appreciate it!
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    (Original post by B_9710)
    Multiply the coefficient of x^2 term by get constant term. In other words multiply a by c to get ac. Then find 2 numbers that multiply to make ac but add to make b - the coefficient of the x term. Then split the x term into whatever these 2 numbers are. There should now be 4 terms in the expression. Now factorise the first 2 terms and then factorise the last 2 terms separately. You should find that there are 2 identical brackets. Then take out the expression in the brackets as a common factor and you should have factorised the quadratic into 2 linear factors.
    I will give an example of this.
    If we have 2x^2 + 9x + 10
    We need to find 2 numbers that multiply to make 2*10 , 20 and add together to make 9 so we can see that the numbers 4 and 5 make this work.
    Now we split the x term - 9x into the 2 numbers that we just found.
    So 9x = 4x+5x
    So the quadratic becomes
    2x^2 + 4x+ 5x + 10
    We factorise 2x^2 + 4x and 5x+10 separately
    So we get 2x(x+2) +5(x+2)
    So you see we have 2 identical brackets (x+2) common to both the 2x term and the constant term 5
    Now if we take out (x+2) as a common factor we get (x+2)(2x+5)
    So now we see that
    2x^2 + 9x + 10 = (2x+5)(x+2)
    Totally get it now. Thanks so much!!
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    (Original post by YsfAli)
    That actually makes more sense to me as my teacher doesn't really bother explaining little things in detail. Thanks a lot really do appreciate it!
    No problem!
 
 
 
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