Okay, so I have a paper from math that involves factoring polynomials, which I have no idea how to do! I'll post the 5 sample questions I have, and you do NOT have to answer all of them. Why, even answering just a couple would be helpful!! What I really need is someone to take me step-by-step to help me figure out how to solve these problems. Please make your explanations thorough! Remember- quality, not quantity. But both would be nice. (: Here are the problems. Any you can answer would be awesome- 10 points for best answer!
p^2-169q^2
3x^2-19x+6
x^2-3x-10
5m^4+70m^3+254m^2
198q^3r^2-184q^2r^2+18qr^2
Also, if you are smart enough to factor the last two...kudos to you, and I love you forever. XD Thanks guys!
p^2-169q^2
3x^2-19x+6
x^2-3x-10
5m^4+70m^3+254m^2
198q^3r^2-184q^2r^2+18qr^2
Also, if you are smart enough to factor the last two...kudos to you, and I love you forever. XD Thanks guys!
-
p^2-169q^2
know that a^2 - b^2 = (a+b)(a-b)
This makes me suspect that 169 is a square number, and it turns out 169 = 13^2
p^2 - 13^2q^2
(p + 13q)(p - 13q)
3x^2-19x+6 we know that it will factor something like (3x )(x )
the negative 19 x tells us at least one of the missing numbers is negative and the positive 6 tells us the missing numbers have the same sign. Thus both missing numbers are negative so we have
(3x - )(x - ) now we know 6 factors to 1 and 6, it also factors to 2 and 3. Since 19 is much larger than 3 and 6 we probably want to multiply the 3x by the larger number.
(3x- 2)(x - 3) gives us -9x and -2x which adds to -11x so we know that's wrong.
(3x -1)(x -6) which gives us -18x - 1x adding to -19x
x^2-3x-10 because 10 is negative we know our factorization will have one positive and one negative number so we have
(x+ )(x- ) because 3x is negative we know the larger number will be negative. Now 10 factors to 10 and 1 or 2 and 5. 10-1= 9, and 5-2=3 so we are going to use 5 and 2.
(x+2)(x-5)
5m^4+70m^3+254m^2 first factor out m^2
m^2(5m^2 +70m +254) attempting to divide 254 by all prime numbers below 20 shows that the only ways to factor it are 254 and 1 or 127 and 2, neither of which will factor in our equation so the simplest form is m^2(5m^2 +70m +254)
198q^3r^2-184q^2r^2+18qr^2 first factor out r^2
r^2(198q^3 - 184q^2 + 18q) next we see we can factor out a q
qr^2 (198q^2 - 184q +18) You'll notice that all the terms are even so we can factor out a 2
2qr^2 (99q^2 -92q +9) we know that 9 is only divisible by 3 and when we check the other terms we discover 92 is not divisible by 3 (92 -> 9+2 = 11, 11 does not divide evenly by 3 so 92 does not either) so the simplest form is 2qr^2 (99q^2 - 92q + 9)
know that a^2 - b^2 = (a+b)(a-b)
This makes me suspect that 169 is a square number, and it turns out 169 = 13^2
p^2 - 13^2q^2
(p + 13q)(p - 13q)
3x^2-19x+6 we know that it will factor something like (3x )(x )
the negative 19 x tells us at least one of the missing numbers is negative and the positive 6 tells us the missing numbers have the same sign. Thus both missing numbers are negative so we have
(3x - )(x - ) now we know 6 factors to 1 and 6, it also factors to 2 and 3. Since 19 is much larger than 3 and 6 we probably want to multiply the 3x by the larger number.
(3x- 2)(x - 3) gives us -9x and -2x which adds to -11x so we know that's wrong.
(3x -1)(x -6) which gives us -18x - 1x adding to -19x
x^2-3x-10 because 10 is negative we know our factorization will have one positive and one negative number so we have
(x+ )(x- ) because 3x is negative we know the larger number will be negative. Now 10 factors to 10 and 1 or 2 and 5. 10-1= 9, and 5-2=3 so we are going to use 5 and 2.
(x+2)(x-5)
5m^4+70m^3+254m^2 first factor out m^2
m^2(5m^2 +70m +254) attempting to divide 254 by all prime numbers below 20 shows that the only ways to factor it are 254 and 1 or 127 and 2, neither of which will factor in our equation so the simplest form is m^2(5m^2 +70m +254)
198q^3r^2-184q^2r^2+18qr^2 first factor out r^2
r^2(198q^3 - 184q^2 + 18q) next we see we can factor out a q
qr^2 (198q^2 - 184q +18) You'll notice that all the terms are even so we can factor out a 2
2qr^2 (99q^2 -92q +9) we know that 9 is only divisible by 3 and when we check the other terms we discover 92 is not divisible by 3 (92 -> 9+2 = 11, 11 does not divide evenly by 3 so 92 does not either) so the simplest form is 2qr^2 (99q^2 - 92q + 9)
-
p^2-169q^2
=(p-13q)(p+13q)
3x^2-19x+6
= (3x-1)(x-6)
x^2-3x-10
=(x-5)(x+2)
5m^4+70m^3+254m^2
= m²(5m²+70m+254)
198q^3r^2-184q^2r^2+18qr^2
= 2qr²(99q² -62q +9)
=(p-13q)(p+13q)
3x^2-19x+6
= (3x-1)(x-6)
x^2-3x-10
=(x-5)(x+2)
5m^4+70m^3+254m^2
= m²(5m²+70m+254)
198q^3r^2-184q^2r^2+18qr^2
= 2qr²(99q² -62q +9)