how do I solve for x in the following equation: (x-1)(5x+3)?
2007-07-20 13:13:56 · 9 answers · asked by jumba 1 in Science & Mathematics ➔ Mathematics
with this you want to FOIL (first,outer,innner,last)
FIRST: x times 5x=5x^2(5x squared)
OUTER: x times 3=3x
INNER:-1 times 5x=-5x
LAST:-1 times 3= -3
so, the equation will look like
5x^2+3x+(-5x)+(-3)
5x^2 - 2x - 3 <----thats the answer
ps- never set these paratheses to zero UNLESS the problem is equal to zero solve for x.
i hope this helped, these were always my favorite math problems.
2007-07-20 13:30:40 · answer #1 · answered by B A 2 · 1⤊ 0⤋
Hey there!
(x-1)(5x+3) is an expression, not an equation. It would be an equation if it was set equal to a constant, a variable or another expression. I'll assume the expression was set equal to 0.
(x-1)(5x+3)=0 --> Write the problem.
x-1=0 or 5x+3=0 --> Use the zero-product property i.e. if pq=0, then p=0 or q=0.
x=1 or x=-3/5 Solve each equation for x.
So the answer is x=1 or x=-3/5 or {-3/5,1}.
Hope it helps!
2007-07-20 13:22:47 · answer #2 · answered by ? 6 · 0⤊ 0⤋
(x-1)(5x+3) is not an equation, it is an expression.
To solve for the zeros (when the expression = 0), solve
x-1 = 0 or 5x+3 = 0
x = 1 or x = -3/5
2007-07-20 13:19:14 · answer #3 · answered by MooseBoys 6 · 0⤊ 0⤋
An equation requires = sign so this is not an equation. It is an expression.
(x - 1) (5x + 3) = 5x² + 3x - 5x - 3 = 5x² - 2x - 3
2007-07-20 21:41:23 · answer #4 · answered by Como 7 · 1⤊ 0⤋
we have (x-1)(5x+3)=0
soEither (x-1)=0
Or 5x+3=0
Or Either x=1 Or x=-3/5 Ans
2007-07-20 13:28:11 · answer #5 · answered by MAHAANIM07 4 · 0⤊ 0⤋
The equation is not set to be equal to anything. I could expand the equation to be 5x^2 - 2x - 3
2007-07-20 13:25:25 · answer #6 · answered by Anonymous · 1⤊ 0⤋
(x-1)(5x+3)=?
If it is (x-1)(5x+3)=0,then the solution is being given
As we know,the product of two or more numbers cannot be zero unless at lest one of them is zero
Therefore,either x-1=0 or 5x+3=0
If x-1=0,then x=1
If 5x+3=0,then 5x= -3 or x= -3/5
Hence x=1 or -3/5 ans
2007-07-20 13:20:04 · answer #7 · answered by alpha 7 · 0⤊ 0⤋
get x alone.
(x-1)(5x+3)=0
disrtibute
FOIL= first, outside inside last
F 3x^2
O 3x
I -3x
L -3
add
3x^2-3
sub 3
3x^2=3
Divide by 3
x^2=1
x=1
(1-1)(5(1)+3)
0*8=0
understand?
2007-07-20 13:21:50 · answer #8 · answered by notapandabear13 2 · 0⤊ 0⤋
the distributive property! :]
2007-07-20 13:23:34 · answer #9 · answered by Anonymous · 0⤊ 1⤋