tobycutie said:
I was having trouble working out this problem, if u could tell me if i was doing it right or what I was doing wrong it would be greatly appreciated. Thanks!!
Simplify:
(3x^2-x) + (2x-8) + (-x^2+2x-27)
(2x^2-x-21) (2x^2-x-21) (2x^2-x-21)
(2x^2+3x-35) This is what I simplified it to
(2x^2-x-21)
(2x+6)(x-7) And this is then what I factored it to
(2x+6)(x-7)
Then the answer I got was 1 but it the back of my textbook it said the answer was (x+5)
(x+3)
Please Tell me what I am doing wrong
Your difficulty is in the factoring. (2x[sup:1gy1mo86]2[/sup:1gy1mo86] + 3x - 35) and (2x[sup:1gy1mo86]2[/sup:1gy1mo86] - x - 21) are two DIFFERENT and unequal expressions, so they both can't be factored the same way.
Neither of your factorizations are correct (and you should be able to see that, if you multiply (2x + 6)(x - 7)).
I'll help you factor the numerator, using a method I like.
2x[sup:1gy1mo86]2[/sup:1gy1mo86] + 3x - 35
Start by multiplying the coefficient of the x[sup:1gy1mo86]2[/sup:1gy1mo86] term by the constant term: 2*(-35) = -70.
Now, look for two numbers whose product is -70, and whose sum is the coefficient of the middle term, +3. -7 and 10 will work, because -7(10) = -70 and -7 + 10 = 3. Rewrite the middle term as -7x + 10x:
2x[sup:1gy1mo86]2[/sup:1gy1mo86] - 7x + 10x - 35
Factor by grouping. The first two terms have a common factor of x, and the last two terms have a common factor of 5. Remove the common factor from each pair of terms:
x(2x - 7) + 5(2x - 7)
And there's a common factor of (2x - 7). Remove it to get
(2x - 7)(x + 5)
That's the factorization of the numerator.
Try the same process to factor the denominator. Once you have factored both numerator and denominator correctly, you can look for and divide out any factors the numerator and denominator have in common.
