I had a rough day and slept through this class today but I need to do this homework. How do I factor:
a^2 - 3ab - 130b^2
and
ab + 20a^2b - 12b
and how do I solve:
12x^2 + 11x - 15 = 0
and
12x^2 - 36x + 15 = 0
Please break down each step by step. Thanks in advance!
a^2 - 3ab - 130b^2
and
ab + 20a^2b - 12b
and how do I solve:
12x^2 + 11x - 15 = 0
and
12x^2 - 36x + 15 = 0
Please break down each step by step. Thanks in advance!
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Find factors of 130 which is 13x5x2, now construct two out of these such that the difference is 3 . clearly 10 and 13. Now you have the result
(a+10b)(a-13b)
Next
b common removing it you get 20a^2 + a -12 . The result will be (?a+?)(?a-?) to find see ? with a when multiplied should give 20 and ? without a should give -12 also cross multiplication and addition of ? should be 1 to get single a.
5x4 3x4 are the required figures.
so your result is b (5a+4)(4a-3)
15 has only factors 3 and 5 so for solutions (?x-3)(?x+5) or (?x-3)(?x-5) are for the next two.
both are 12x^2 so 2x6 or 3x4 are how it can form.
you need +11 and -36 in cross multiplication and addition
so factors1 is (4x-3)(3x+5) = 0 therefore x=3/4 or -5/3
factors2 is (6x-3)(2x-5) = 0 therefore x=3/6 or 5/2
(a+10b)(a-13b)
Next
b common removing it you get 20a^2 + a -12 . The result will be (?a+?)(?a-?) to find see ? with a when multiplied should give 20 and ? without a should give -12 also cross multiplication and addition of ? should be 1 to get single a.
5x4 3x4 are the required figures.
so your result is b (5a+4)(4a-3)
15 has only factors 3 and 5 so for solutions (?x-3)(?x+5) or (?x-3)(?x-5) are for the next two.
both are 12x^2 so 2x6 or 3x4 are how it can form.
you need +11 and -36 in cross multiplication and addition
so factors1 is (4x-3)(3x+5) = 0 therefore x=3/4 or -5/3
factors2 is (6x-3)(2x-5) = 0 therefore x=3/6 or 5/2
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The first trinomial is just an easy factoring if you treat the terms with "b" as just a number; notice that the coefficient in front of a^2 is 1 and its product -130b^2, so its factors are 10b and -13b which adds to the middle term of "-3b" and -130b^2 is its product. Thus we can factor it as
a^2 - 3ab-130b^2 = a^2 + 10ab - 13ab - 130 b^2 = a(a+10b) -13(a +10b) = (a-13b)(a+10b)
The second trinomial is just about the same, but after factoring it, write each factor equal to zero and solve for the variables.
12x^2+11x-15=(3x+5)(4x-3)=0
so 3x+5=0 and 4x-3=0 which implies x=-5/3 and x=3/4.
Likewise, with the second trinomial factor out a greatest common factor of 3, which gives 3( 4x^2 - 12x+5)=0. To factor from here is straight forward, you get 3(2x-1)(2x-5), set each factor equal to zero and you get 2x-1=0 and 2x-5=0, which implies x=1/2 and x=5/2. Notice the scaling factor of 3 does not affect the x-intercepts.
a^2 - 3ab-130b^2 = a^2 + 10ab - 13ab - 130 b^2 = a(a+10b) -13(a +10b) = (a-13b)(a+10b)
The second trinomial is just about the same, but after factoring it, write each factor equal to zero and solve for the variables.
12x^2+11x-15=(3x+5)(4x-3)=0
so 3x+5=0 and 4x-3=0 which implies x=-5/3 and x=3/4.
Likewise, with the second trinomial factor out a greatest common factor of 3, which gives 3( 4x^2 - 12x+5)=0. To factor from here is straight forward, you get 3(2x-1)(2x-5), set each factor equal to zero and you get 2x-1=0 and 2x-5=0, which implies x=1/2 and x=5/2. Notice the scaling factor of 3 does not affect the x-intercepts.
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Solve:
12x^2 + 11x - 15 = 0
(4x - 3) (3x + 5) = 0
4x - 3 = 0 then x = 3/4
3x + 5 = 0 then x = - 5/3
12x^2 - 36x + 15 = 0 (Factor Out GCF=3)
3(4x^2 - 12x + 5) = 0
3(2x - 5) (2x - 1) = 0
x = 5/2 and x = 1/2
Hope it helps!
12x^2 + 11x - 15 = 0
(4x - 3) (3x + 5) = 0
4x - 3 = 0 then x = 3/4
3x + 5 = 0 then x = - 5/3
12x^2 - 36x + 15 = 0 (Factor Out GCF=3)
3(4x^2 - 12x + 5) = 0
3(2x - 5) (2x - 1) = 0
x = 5/2 and x = 1/2
Hope it helps!
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a^2 - 3ab - 130b^2:
First find multiples of -130 that add to get -3. That would be -10 and 3.
Next write like this:
(a-10)(a+3)
First find multiples of -130 that add to get -3. That would be -10 and 3.
Next write like this:
(a-10)(a+3)