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Help solve stupid math problem (1 Viewer)
- Thread starter Brodie28
- Start date
Use the factor theorem and the fact that the factors are going to be negative (or complex) since all the coefficients > 0.
take (x+1)(3x³+8x²+7x+2) and consider 3x³+8x²+7x+2 by itself
when x=-1 ------> 3x³+8x²+7x+2 = -3 +8 -7 +2 = 0
hence 3x³+8x²+7x+2 = (x+1)(3x<sup>2</sup>+5x+2) = (x+1)(3x+2)(x+1)
therefore (x+1)(3x³+8x²+7x+2) = (x+1)(x+1)(3x+2)(x+1) = (x+1)<sup>3</sup>(3x+2)
EDIT:
Basically, subbing in values will tell you what the factors are:
If you have a polynomial P(x) = a<sub>n</sub>x<sup>n</sup> + a<sub>n-1</sub>x<sup>n-1</sup>+ ...+ a<sub>2</sub>x<sup>2</sup> + a<sub>1</sub>x + a<sub>0</sub>
and P(a) = 0 then (x-a) is a factor of P(x)
Using this you know that if you sub in a value 'a' for x and the polynomial reduces to zero then you have found a factor
---> (x-a) which is a big help in factorisation.
take (x+1)(3x³+8x²+7x+2) and consider 3x³+8x²+7x+2 by itself
when x=-1 ------> 3x³+8x²+7x+2 = -3 +8 -7 +2 = 0
hence 3x³+8x²+7x+2 = (x+1)(3x<sup>2</sup>+5x+2) = (x+1)(3x+2)(x+1)
therefore (x+1)(3x³+8x²+7x+2) = (x+1)(x+1)(3x+2)(x+1) = (x+1)<sup>3</sup>(3x+2)
EDIT:
Basically, subbing in values will tell you what the factors are:
If you have a polynomial P(x) = a<sub>n</sub>x<sup>n</sup> + a<sub>n-1</sub>x<sup>n-1</sup>+ ...+ a<sub>2</sub>x<sup>2</sup> + a<sub>1</sub>x + a<sub>0</sub>
and P(a) = 0 then (x-a) is a factor of P(x)
Using this you know that if you sub in a value 'a' for x and the polynomial reduces to zero then you have found a factor
---> (x-a) which is a big help in factorisation.
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