Can you factor these for me? Please show all steps!?
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x2x6=0 (The 2 is an exponent)
x2+5x+4=0 (The 2 is an exponent)
x24x5=0 (The 2 is an exponent, they're all going to be that way so you get the point)
x2+x12=0 ...show more
Best Answer
Hi,
For problems starting with x², look at the last sign. If it is positive, find factors of the constant (the last number) that add to the middle number. If the last sign is negative, find
factors of the constant (the last number) that subtract to the middle number.
x2x6=0
(x  3)(x + 2) = 0
x = 3 or 2
x2+5x+4=0
(x + 4)(x + 1) = 0
x = 4 or 1
x24x5=0
(x  5)(x + 1) = 0
x = 5 or x = 1
x2+x12=0
(x + 4)(x  3) = 0
x = 4 or x = 3
x23x10=0
(x  5)(x + 2) = 0
x = 5 or x = 2
x2+7x+12=0
(x + 4)(x + 3) = 0
x = 4 or x = 3
x2+2x15=0
(x + 5)(x  3) = 0
x = 5 or x = 3
x29x+14=0
(x  7)(x  2) = 0
x = 7 or x = 2
x27x18=0
(x  9)(x + 2) = 0
x = 9 or x = 2
x2+4x45=0
(x + 9)(x  5) = 0
x = 9 or x = 5
x2+9x+18=0
(x + 6)(x + 3) = 0
x = 6 or x = 3
x28x+15=0
(x  5)(x  3) = 0
x = 5 or x = 3
x2+6x+5=0
(x + 5)(x + 1) = 0
x = 5 or x = 1
I hope that helps!! :)
For problems starting with x², look at the last sign. If it is positive, find factors of the constant (the last number) that add to the middle number. If the last sign is negative, find
factors of the constant (the last number) that subtract to the middle number.
x2x6=0
(x  3)(x + 2) = 0
x = 3 or 2
x2+5x+4=0
(x + 4)(x + 1) = 0
x = 4 or 1
x24x5=0
(x  5)(x + 1) = 0
x = 5 or x = 1
x2+x12=0
(x + 4)(x  3) = 0
x = 4 or x = 3
x23x10=0
(x  5)(x + 2) = 0
x = 5 or x = 2
x2+7x+12=0
(x + 4)(x + 3) = 0
x = 4 or x = 3
x2+2x15=0
(x + 5)(x  3) = 0
x = 5 or x = 3
x29x+14=0
(x  7)(x  2) = 0
x = 7 or x = 2
x27x18=0
(x  9)(x + 2) = 0
x = 9 or x = 2
x2+4x45=0
(x + 9)(x  5) = 0
x = 9 or x = 5
x2+9x+18=0
(x + 6)(x + 3) = 0
x = 6 or x = 3
x28x+15=0
(x  5)(x  3) = 0
x = 5 or x = 3
x2+6x+5=0
(x + 5)(x + 1) = 0
x = 5 or x = 1
I hope that helps!! :)
Other Answers (8)
Rated Highest
too many!

seriously...all of these

wow if it was only one or 2 its okay but theres soo many. btw just by looking at them, they arent that hard to factor, just too timeconsuming

You are being asked questions about them and you are NOT PREPARED !
These are basic and simple trinomial factoring problems. You should be able to do these. If you really can't do any of these, then you must go to your teacher. Obviously, the concept of using your text book is not in your repertoire !
You will have to pass tests on this material. I suggest you learn it !
QED 
I will teach you how to do this. The answers are in the back of your text book.
eg. x^2 + 5x +6 = 0
( x )( x ) = 0 First what to multiply to get x^2 x * x = x^2
( x + 3 )( x +2 ) = 0 What two numbers have a product of 6 but add to 5.
Answer 3 and 2
Therefore x+3 =0 or x+2 = 0 So x = 3 or 2
x^2  3x + 10 = 0
( x  5 )( x +2 ) =0 x * x = x^2 , 5 * 2 = 10 , 5 +2 =3
Therefore x5 =0 or x+2 = 0
so x = 5 or 2 
(x3)(x+2)
(x+1)(x+4)
(x5)(x+1)
(x+4)(x3)
(x5)(x+2)
(x+4)(x+3)
(x+5)(x3)
(x7)(x2)
(x9)(x+2)
(x+9)(x5)
(x+3)(x+6)
(x3)(x5)
(x+1)(x+5) 
x2+5x+4=0=(x+4)(x+1)
x24x5=0=(x5)(x+1)
x2+x12=0 = (x+4)(x3)
x23x10=0=(x5)(x+2)
x^2 + x 12 = (x + 4)(x3)
x^2 9x +14 = (x7)(x2)
x27x18=0 = (x9)(x+2)
x2+4x45=0 = (x+9)(x5)
x2+9x+18=0 = (x+3)(x+6)
x28x+15=0 = (x3)(x_5)
x2+6x+5=0 = (x+5)(x+1) 
x^2  x  6 = 0 Factor (x +2)(x  3)
x^2 + 5x + 4 = 0 Factor (x + 4)(x + 1)
x^2 + x  12 = 0 Factor (x + 4)(x  3)
x^2  3x  10 = 0 Factor (x  5)(x + 2)
x^2 + 7x + 12 = 0 Factor (x+3)(x + 4)
x^2 + 2x  15 = 0 Factor (x + 7)(x  5)
x^2  9x + 14 = 0 Factor (x  7)(x  2)
x^2  7x 18 = 0 Factor (x  9)(x + 2)
x^2 + 4x  45 = 0 Factor (x + 9)(x  5)
x^2 + 9x + 18 = 0 Factor (x = 6)(x + 3)
x^2 8x + 15 = 0 Factor (x  5)(x  3)
x^2 + 6x + 5 = 0 Factor (x = 5)(x + 1)
OK all done
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