I'm trying to factor the following equation:
4x(squared) + 19x - 5
I seem to be a complete idiot when it comes to math because I know I've done this sort of problem before, but I cannot for the life of me remember how to do it. Thanks for your help.
How do I factor this equation?
2009-04-10 2:45 pm
回答 (10)
2009-04-10 2:49 pm
✔ 最佳答案
DecompositionMultiply the first terms coefficient (4) and the last terms coefficient (-5), then find all the factors. One pair of factors will add up to give 19X.
(4)(-5) = -20
Factors of -20:
1,-20
-1,20
2,-10
-2,10
4,-5
-4,5
4X^2 + 19X - 5
4X^2 + 20X - X - 5
4X(X+5) - 1(x+5)
(4X-1)(X+5)
X = 1/4
X = -5
2009-04-10 2:53 pm
4x² + 19x - 5 = 0
x² + 19/8x = 5/4 + (19/8)²
x² + 19/8x = 80/64 + 361/64
(x + 19/8)² = 441/64
x + 19/8 = 21/8
= x + 19/8 - 21/8, = x - 2/8, = x - 1/4, = 4x - 1
= x + 19/8 + 21/8, = x + 40/8, = x + 5
Answer: (4x - 1)(x + 5)
x² + 19/8x = 5/4 + (19/8)²
x² + 19/8x = 80/64 + 361/64
(x + 19/8)² = 441/64
x + 19/8 = 21/8
= x + 19/8 - 21/8, = x - 2/8, = x - 1/4, = 4x - 1
= x + 19/8 + 21/8, = x + 40/8, = x + 5
Answer: (4x - 1)(x + 5)
2016-12-08 11:46 pm
factoring is simplifying a equation... x²+8x+12 so which you may take the factors of 12: a million,12 2,6 3,4 then you definitely discover the pair that provides to furnish you 8. subsequently its 2 and six. so that then you definitely write (x+6)(x+2)
2009-04-10 6:22 pm
4x^2 + 19x - 5 =
4x^2 + 20x - x - 5 =
4x(x + 5) - (x + 5) =
(4x - 1)(x + 5)
4x^2 + 20x - x - 5 =
4x(x + 5) - (x + 5) =
(4x - 1)(x + 5)
2009-04-10 4:29 pm
4x^2 + 19x - 5
= 4x^2 + 20x - x - 5
= (4x^2 + 20x) - (x + 5)
= 4x(x + 5) - 1(x + 5)
= (x + 5)(4x - 1)
= 4x^2 + 20x - x - 5
= (4x^2 + 20x) - (x + 5)
= 4x(x + 5) - 1(x + 5)
= (x + 5)(4x - 1)
2009-04-10 2:57 pm
4x^2+19x-5=0
(4x-1)(x+5)=0
4x-1=0 x+5=0
4x=1...x=-5
x=1/4
so x={1/4,-5} answer//
(4x-1)(x+5)=0
4x-1=0 x+5=0
4x=1...x=-5
x=1/4
so x={1/4,-5} answer//
2009-04-10 2:53 pm
4x² + 19x - 5
= (4x - 1)(x + 5)
Done
= (4x - 1)(x + 5)
Done
2009-04-10 2:51 pm
(4x-1)(x+5)
2009-04-10 2:51 pm
4x^2+19x-5
(4x-1)(x+5)
(4x-1)(x+5)
2009-04-10 2:50 pm
If you don't recognize a solution right away, you list the possibilities, using the factors of the square term and the constant term. In other words, suppose the trinomial is of the form ax^2+bx+c. If it can be factored, the solution will look like (dx+e)(fx+g).
We know that a=df and c=eg, so we can set up the possibilities:
the first terms of the factors are either 4x and x or 2x and 2x. The second terms of the factors are either +5 and -1, or -5 and +1.
So the possibilities are:
(4x+5)(x-1)
(4x-1)(x+5)
(4x-5)(x+1)
(4x+1)(x-5)
(2x+5)(2x-1)
(2x-5)(2x+1)
By testing each one, you reach the solution: (4x-1)(x+5)
Of course the best way is to recognize the pattern, but that just comes with practice.
We know that a=df and c=eg, so we can set up the possibilities:
the first terms of the factors are either 4x and x or 2x and 2x. The second terms of the factors are either +5 and -1, or -5 and +1.
So the possibilities are:
(4x+5)(x-1)
(4x-1)(x+5)
(4x-5)(x+1)
(4x+1)(x-5)
(2x+5)(2x-1)
(2x-5)(2x+1)
By testing each one, you reach the solution: (4x-1)(x+5)
Of course the best way is to recognize the pattern, but that just comes with practice.
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