factor the expression

## how do you factor the expression 4x²+29x+30

# 6 Answers

To factor ax^{2} + 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

4x^{2} + 29x + 30 = 4x^{2} + 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 6x^{2}+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)

# Comments

I meant (3x-2)(2x+5), should have proof read this before posting

This is a good method, similar to what I use, and students find it easier to understand.

Thank you, students I work with also find it easier since they're doing almost the exact same steps they would do if the leading coefficient were 1

ax^{2} + bx + c = a(x - x_{1})(x - x_{2}) ,

x_{12} are roots of equation ax^{2} + bx + c = 0

- b ± √(b^{2} - 4ac)

x_{12} = —————————

2a

**~~~~~~~~~~~~~ **

4x^{2} + 29 + 30 = 0

-29 ± √(29^{2} - 4*4*30)

x_{12} = ————————————

2*4

- 29 ± 19

x_{12} = ——————

8

x_{1} = - 48/8 = - 6

x_{2} = - 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

next

(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)

(4x+3)(x+10)

(4x+6)(x+5) ...

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.

## Comments

Note: to understand why the above method of factoring works, consider the following ---

write the general Quadratic Trinomial, in factored form, as follows:

Ax^2 + Bx + C = (ax + b)(cx + d), where A, B, and C are given (for example, A = 4; B = 29; C = 30) and where a, b, c, and d are to be determined

then, using the Distributive Property (i.e., FOIL), write the following:

Ax^2 + Bx + C = (ac)x^2 + (ad + bc)x + bd, implying A = ac; B = ad + bc; C = bd;

now, note that: AC = (ac)(bd) = (ad)(bc);

thus, we're looking for two numbers, "ad" and "bc", whose product is

(ad)(bc) = AC, and whose sum is B = ad + bc

the remainder of the argument then follows, as in the example above

one further note --- if it turns out that we can find no numbers a, b, c, d such that

(ac)(bd) = (ad)(bc) = AC, and (ad + bc) = B,

then we conclude that Ax^2 + Bx + C is prime, i.e., non-factorable (at least, non-factorable "over the integers", i.e., with each of a, b, c, d being integers).

one last note --- from the previous note, we see that one example of a non-factorable Quadratic Trinomial would be the following: 4x^2 + 31x + 30, as would any Quadratic Trinomial of the form 4x^2 + Bx + 30, where B is any of the numbers {30, 31, 32, 33, 34, 35, 36, 37, 38, etc., etc., etc., ....} --- note that if, in the above list, B is even (a multiple of 2), then a factor of 2 could be factored out from 4x^2 + Bx + 30, but apart from that, 4x^2 + Bx + 30 would be non-factorable, i.e., it would not be factorable as a product of two linear factors, each of the form (ax + b)

a related question for teachers of the above topic could be the following:

to continue the previously interrupted comment:

a related question for teachers of the above topic could be the following:

for what integer values of "B" is 4x^2 + Bx + 30 factorable?

Comment