Quadratic trinomial problem

[gcooper]

4x^2 +5x +3

So, the old: product must be 3, and addition must be equal to 5 method doesn't work here(WHY?).

And I can't really figure another solution to this.
I have a book that says: list all pairs of factors of 4x^2, then list all pairs of factors of 3, and then test all possible factors of 4x^2 and 3 to find out which factor produced the correct 5x.

I can't comprehend how to use the above solution.

The factors of 4x^2 are 1 x 4x and 2x x 2x
The factors of 3 are 1 and 3.

I tried setting up something like (1x+1)(2x+3) but obviously that is nowhere near the correct answer.

So can someone explain exactly how that advice from the book works?

 

[Ishuda]

4x^2 +5x +3

So, the old: product must be 3, and addition must be equal to 5 method doesn't work here(WHY?).

And I can't really figure another solution to this.
I have a book that says: list all pairs of factors of 4x^2, then list all pairs of factors of 3, and then test all possible factors of 4x^2 and 3 to find out which factor produced the correct 5x.

I can't comprehend how to use the above solution.

The factors of 4x^2 are 1x x 4x and 2x x 2x
The factors of 3 are 1 and 3.

I tried setting up something like (1x+1)(2x+3) but obviously that is nowhere near the correct answer.

So can someone explain exactly how that advice from the book works?

Look at the middle term of 4x² +5x +3. Since that term contains an x, both factors of 4x² which we are going to use must contain an x, i.e. 1x and 4x or 2x and 2x as you wrote (but see corrected typo above in red). The factors of 3 are, as you mentioned 1 and 3.

Working with the 1x and 4x factors first we choose the first combination as 1 with 1x and thus 3 with 4x or (1 x + 1) (4 x + 3); sum of cross terms (4x X 1 + 1x X 3) is 7x, doesn't work.

The second combination is then 1 with 4x and thus 3 with 1x or (4x + 1) (x + 3); sum of cross terms is 13 x, doesn't work.

So we have to go to the 2x and 2x factors and the third combination is chosen as 1 with 2x and thus 3 with 2x or (2x + 1) (2x + 3); sum of cross terms is 5x, that works and we can stop.

The more you do of these the better you will get at 'guessing' the proper solution the first time around.

 

[gcooper]

I'm sorry its quite late here and I might look at the problem in a wrong way, but:

(2x + 1) (2x + 3)

doesn't that equal: 4x^2 + 3 + 6x + 2x thus 4x^2 + 8x + 3 ?

 

[Ishuda]

I'm sorry its quite late here and I might look at the problem in a wrong way, but:

(2x + 1) (2x + 3)

doesn't that equal: 4x^2 + 3 + 6x + 2x thus 4x^2 + 8x + 3 ?

Of course you are correct - it was sleepy out here too and I'm sorry I missed that. What we have shown then is that you can not factor the expression with integer expressions. Actually what the book should say is, "If the equation can be expressed as two linear expressions with integer coefficients, then list all pairs of factors of ...".

There is a theorem which says if
f(x) = a_n xⁿ + a_n-1 x^n-1 +... + a₁ x + a₀
then we can write
f(x) = a_n (x-x_n) (x-x_n-1) (x-x_n-2)...(x-x₁) (x-x₀)
where the x_j are the roots of f and some of them may be the same.

In this case
f(x) = 4x² +5x +3
and the roots of f are, by the quadratic formula,
\(\displaystyle \frac{-5\pm\sqrt{-23}}{8}\)
so there are no real valued roots of f let alone rational roots.

 

[gcooper]

thanks for the answer. I guess it's actually my fault. The book says to actually write "prime" if I can't manage to factor it the way it describes.

 

[HallsofIvy]

(x- a)(x- b)= x^2- (a+ b)x+ ab. I guess the "product must be 3, and addition must be equal to 5" refer to ab and a+ b.

But your quadratic, \(\displaystyle 4x^2+ 5x+ 3\), has leading coefficient "4", not "1" so that does not apply:
(ax- b)(cx- d)= acx^2- (ad+ bc)x+ ad is much more complicated!

One way to check whether or not a quadratic can be factored, using integer coefficients, is to use the discriminant from the quadratic formula.
If the quadratic \(\displaystyle ax^2+ bx+ c\) can be factored as \(\displaystyle a(x- x_1)(x- x_2)\) then \(\displaystyle x_1\) and \(\displaystyle x_2\) are roots of the equation and so must be of the form \(\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}\). The "discriminant" is \(\displaystyle b^2- 4ac\) which, here, is \(\displaystyle 5^2- 4(4)(3)= 25- 49= -23\). That is, the roots are not "real numbers", much less integers, so we cannot factor the quadratic in terms of integer coefficients.