2007-01-17 09:21:51 · 9 answers · asked by brunettebelle10 2 in Science & Mathematics ➔ Mathematics
This equation is in the form of a x ^2 + bx + c.
First start with c. 5 is a prime number, therefore it is only divisible by 5 and 1.
Next start with a. 3 is a prime number, therefore it is only divisible by 3 and 1.
a, b, and c are all > 0 therefore we need a combination of the roots of a and the roots of b to be multiplied and then added together to equal b or 16.
Now, 3 x 5 = 15 and 1 x 1 = 1 and 15 + 1 = 16.
Therefore, the factors of this equation are
(3x + 1)(x + 5).
To take it a step further the solution of this equation is
x = -1/3 or -5.
Hope this helps.
2007-01-17 09:32:13 · answer #1 · answered by John 2 · 0⤊ 1⤋
Michael Sullivan in his text book 'Algebra and Trigonometry' has a good method for factoring expressions where the coefficient of the first term is more than 1.
Let Ax^2 +Bx +C be the form so that A,B are coefficients and C is the constant.
Multiply A times C and get the product. Here it is 15 (call it AC).
Make a list of all the pairs of numbers which when multiplied together give you AC.
Here that list would be:
15*1
-15*-1
3*5
-3*-5
You have to include pairs of negative numbers when AC is positive because negative times negative produces positive.
Then look at the pairs. Which pair of numbers that has the product AC when ADDED together gives you B?
B is 16.
The only pair here that adds up to 16 is 15 and 1.
Rewrite the original expression:
3x^2 +15x +1x (this is your pair! and it is going to make the middle x term so we're talking here about quantities of xs.) +5.
The expression now reads 3x^2 +15x +1x +5
Now you factor by grouping: that is you try to make two groups of factors by finding common factors, first using the first term and another term and then pulling a common factor out of the group that's left.
You are aiming to end up with a factor that repeats.
In this case, the grouping gives you:
3x(x+5)+1(x+5)
You were looking for a factor that repeats. It is (x+5). The factors that you pulled out (3x and 1) make up a new factor.
Your result is (3x+1) (x+5)
2007-01-21 16:38:18 · answer #2 · answered by kathyw 7 · 0⤊ 0⤋
(3x+1)(x+5) = 3x^2+16x+5
2007-01-17 17:40:22 · answer #3 · answered by hevans1944 5 · 0⤊ 0⤋
(3x + 1)(x+5)
2007-01-17 17:35:14 · answer #4 · answered by db 2 · 0⤊ 0⤋
(3x +1)(x+5)
2007-01-17 17:26:26 · answer #5 · answered by cutest pooky 3 · 0⤊ 0⤋
(x+5)(3x+1)
2007-01-17 17:26:43 · answer #6 · answered by 7 · 0⤊ 0⤋
(3x+1)(x+5)=0
2007-01-17 17:26:03 · answer #7 · answered by VanessaM 3 · 0⤊ 1⤋
(3x+1)*(x+5)
2007-01-17 17:32:41 · answer #8 · answered by matthewjc314 3 · 0⤊ 0⤋
(3x+1)(x+5)
2007-01-17 17:25:29 · answer #9 · answered by Chris C 2 · 0⤊ 1⤋