Factor completely 9x^2 + 42x + 49.
A. (3x + 7)(3x − 7)
B. (9x − 7)(9x − 7)
C. (3x + 7)(3x + 7)
D. (9x + 7)(9x + 7)
HELP PLEASE DON'T UNDERSTAND
since the 1st and last are perfect squares, look for
(3x+7)^2 = 9x^2+42x+49
A. OR C. ?
HUH? I told you which one. See any minus signs anywhere?
oh thank you Steve and monalisa
To factor the given expression, 9x^2 + 42x + 49, we can use the factoring method known as "quadratic trinomial."
First, we look for two numbers that multiply to give us the product of the coefficient of the squared term (which is 9) and the constant term (which is 49), and also add up to give us the coefficient of the middle term (which is 42x).
In this case, there is only one possible combination: 7 and 7. Both 7 and 7 multiply to give us 49, and when we add them, we get 14, which is not the same as 42.
Therefore, it appears this quadratic expression cannot be factored using integer factors. However, we can still check the answer choices to see if any are valid:
A. (3x + 7)(3x − 7):
(3x)(3x) = 9x^2
(3x)(-7) + (7)(3x) = -21x + 21x = 0x
(7)(-7) = -49
So, the answer choice A does not yield the original expression when multiplied out.
B. (9x − 7)(9x − 7):
(9x)(9x) = 81x^2
(9x)(-7) + (-7)(9x) = -63x - 63x = -126x
(-7)(-7) = 49
Again, answer choice B does not yield the original expression when multiplied out.
C. (3x + 7)(3x + 7):
(3x)(3x) = 9x^2
(3x)(7) + (7)(3x) = 21x + 21x = 42x
(7)(7) = 49
Finally, answer choice C does yield the original expression when multiplied out. Therefore, the correct answer is C.