factor the expression
how do you factor the expression 4x²+29x+30
To factor ax2 + bx + c whenever it is possible, just write b as a sum of two integers that have a product ac.
Thus we want m+n = 29 and mn = 4*30 = 120
m and n will be positive and from mn=120 we have the following possibilities.
(1,120), (2,60), (3,40), (4,30), (5,24), (6,20), (8,15), (10,12).
The one with m+n=29 is (5,24).
Thus we have
4x2 + 29x + 30 = 4x2 + 5x + 24x + 30 = x(4x + 5) + 6(4x + 5) = (x + 6)(4x + 5)
Consider: 4(30) = 120
then, consider various ways to multiply two numbers together to get 120:
1 120 note: 1 + 120 = 121
2 60 note: 2 + 60 = 62
3 40 note: 3 + 40 = 43
4 30 note: 4 + 30 = 34
5 24 note: 5 + 24 = 29 Bingo!
then write: 4x^2 + 29x + 30 = 4x^2 + (5x + 24x) + 30 = (4x^2 + 5x) + (24x + 30) =
x(4x + 5) + 6(4x + 5) = (x + 6)(4x + 5) you're done
Roman and Herb's method of factoring by grouping works fine. There is a slight variation to that which can save a few steps.
Once you have determined your 2 numbers simply divide them by the leading coefficient and you can immediately put those number into your factored form.
For example in this problem you've determined 5 and 24 are the 2 numbers we need, dividing these by the leading coefficient of 4 we get 5/4 and 6. Anytime you see a fraction at this point the denominator will be the coefficient in the factored form so we get (4x+5)(x+6) just like everyone else got.
Another example would be 6x2+11x-10.
Multiplying the first last we get -60 so it turns out the 2 numbers that work in this case would be -4 and 15
dividing each by the leading coefficient we get -2/3 and 5/2 so the factored form becomes (3x-2)(5x+2)
ax2 + bx + c = a(x - x1)(x - x2) ,
x12 are roots of equation ax2 + bx + c = 0
- b ± √(b2 - 4ac)
x12 = —————————
4x2 + 29 + 30 = 0
-29 ± √(292 - 4*4*30)
x12 = ————————————
- 29 ± 19
x12 = ——————
x1 = - 48/8 = - 6
x2 = - 10/8 = - 5/4
4 * (x - (-6)) * (x - (- 5/4)) = (x + 6)(4x + 5)
To factor a polymonial that has a number larger than 1 in the quadrtic term:
1) find all possible factors of 4x^2 : (2x)(2x) or (4x)x
2) find all possible factors constant term, 30: (5)(6) , (30)(1), (10)(3) and (-5)(-6) , (-30)(-1) , (-10)(-3)
3) consider the lnear term which must have a sum of 29.
4) By a process of elimunation through trial and error you will fid the correct solution
5) you can elliminate the negative products that produce 30 (in step 2)
6) list possble solution
(2x + 3)(2x + 10)
(2x +5)(2x + 6)
(2x + 1)(2x + 30)
middle term sum
20x + 6x = 26x wrong
12x + 10x = 22x
60x +2x = 62 x
(4x + 30)(x + 1) = 4x + 30x = 34x
(4x + 10)(x + 3) = 12x + 10x = 22x
(4x + 5)(x +6)= 24x + 5x = 29x correct
Answer is (4x + 5)(x + 6) = 4x^2 + 24x + 5x + 30 = 4x^2 + 29x +30
Bear in Mind if you did not find the correct solution after the above trial and error you would contnue :
by switiching the order of the above products, such as:
(4x+ 1)(x +30)
Factoring is essentially, simplifying the expression. Usually it is used to find the roots of an equation. When you have a number in front of the x^2, you should always attempt to factor out that number. Meaning, all the other numbers can be divided by that number. For example: 2x^2+2x+10 could be 2(x^2+2+5). Then one could factor the expression further without the 2. However, in this case, the 4 cannot be factored out.
My next step is to determine the factors of 30.
1,30 2,15 3,30 5,6
Now that we have the factors, we can try and figure out which solution works best.
My first option is (2x+____)(2x+____). In this case, it would not matter which number went in which spot because there are identical. However through trial and error, one can see that not of our factors will finish the solution.
The second option is (4x+__)(x+___). In this case, it does matter which number goes in which spot. The best way to figure this out is through trial and error.
The answer is (4x+5)(x+6).
If we were to expand the above equation we would get:
4x^2+24x+5x+30=4x^2+29x+30. The answer works.
This is how I knew the numbers were 5 and 6 and why they went in each spot.
Again our factor option are: 1,30 2,15 3,10 and 5,6
If we put the 30,15, or 10 in the (x+___) spot, it would be multiplied by 4 and be over 30 so you know that none of those answers could go there.
(4x+30)(x+1)=4x^2+4x+30x+30=4x^2+34x+30. Our solution does not match. The "x" value is too high.
(4x+15)(x+2)=4x^2+8x+15x+30=4x^2+23x+30. Our solution does not match. The "x" value is too low.
(4x+10)(x+3)=4x^2+12x+10x+30=4x^2+22x+30. Our solution does not match. The "x" value is too low.
(4x+6)(x+5)=4x^2+20x+6x+30=4x^2+26x+30. Out solution does not match. The "x" value is too low.
This is why (4x+5)(x+6) is the correct answer.