(1) x^2+3x-28
(2)x^2+14x+40
(3)5y^2+2y-3
(4)9x^2-15x+6
this is like, u have to make a T chart if u know what i mean, but please know what i mean, guys. i am begging u all.............thank u.........
2007-05-26 09:40:58 · 6 answers · asked by soul_dude 2 in Science & Mathematics ➔ Mathematics
Ax^2 + Bx + C = 0
to factor, look for two numbers that multiply to A*C and add up to B
1) x^2 + 3x - 28 = 0
A = 1
B = 3
C = -28
look for two numbers that multiply to 1*-28 = -28 and add up to 3
the two numbers are 7 and -4.
Since the leading coeffecient is 1, plug the two numbers in
(x + 7) (x - 4)
2) Two numbers that multiply to 40 andd add to 14.
The two numbers are 10 and 4
(x + 10) (x + 4)
3) The process is still the same but it takes alitle bit more work.
5y^2 + 2y - 3
A = 5
B = 2
C = -3
look for two numbers that mulitply to 5*-3 = -15 and add to 2
the two numbers are 5 and -3
Notice that the leading coeffecient is NOT 1, you have to factor by groups.
Replace the middle term with the two numbers you just found. (remember to include y behind them)
5y^2 + 5y - 3y - 3
factor by groups
(5y^2 + 5y) + (-3y - 3)
5y(y + 1) + -3(y + 1)
5y(y+1) - 3(y + 1)
(y + 1) (5y - 3)
you try number 4
2007-05-26 09:55:14 · answer #1 · answered by 7 · 0⤊ 1⤋
(1) x^2+3x-28 = ( x +7 )( x - 4 )
(2)x^2+14x+40= ( x +10 )(x + 4 )
(3)5y^2+2y-3 = ( 5y -3)(y + 1)
(4)9x^2-15x+6 = 3 ( 3x^2 - 5 x + 2 )
= 3 ( 3x + 2 )( x + 1 )
2007-05-26 17:34:02 · answer #2 · answered by muhamed a 4 · 0⤊ 0⤋
Question 1
(x + 7).(x - 4)
Question 2
(x + 10).(x + 4)
Question 3
(5x - 3).(x + 1)
Question 4
(3x - 2).(3x - 3)
All above done quickly using trial and error to find factors.
eg in question 1 the only possible factors for x² are x and x so these are fixed. Then try factors for - 28 that add to give 3x. These are 7 and - 4. Hence answer
T chart??
2007-05-27 03:05:05 · answer #3 · answered by Como 7 · 0⤊ 0⤋
(1) Factor x^2+3x-28
First: multiply the 1st & 3rd term to get (-28). find two numbers that give you (-28) when multiplied & 3 (2nd term) when added/subtracted. the numbers are (7 & - 4).
Sec: rewrite the expression with the new middle terms.
x^2 + 7x - 4x - 28
Third: with 4 terms - group "like" terms & factor both sets of parenthesis.
(x^2 + 7x) - (4x - 28)
x(x + 7) - 4(x + 7)
(x + 7)(x - 4)
Follow the same pattern to factor the other expressions.
2007-05-26 20:07:32 · answer #4 · answered by ♪♥Annie♥♪ 6 · 1⤊ 0⤋
1. (x+7)(x-4) x=-7 or 4
2. (x+4)(x+10) x=-10 or -4
3. (5y-3)(y+1) y= 3/5 or -1
4. (3x-3)(3x-2) x= x=1 or 2/3
2007-05-26 19:23:52 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 1⤋
wow mukmin i had the same problem lol
2007-05-27 02:34:24 · answer #6 · answered by janely 1 · 0⤊ 0⤋