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a(4a - 3)(3a - 5)

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13y ago

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(3y - 5)(y + 5)


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3x^(2) +9x - 2x -6 Collect 'like terms'. Hence 3x^(2) + 7x - 6 Next write down all the factors of '3' and '6' Hence 3 ; 1' & 3' 6 ; 1,6 ; 2,3. From these pairs of number we select a pair from each coefficient, that add/multiply to '7' . Hence (3' x 3 ) & (1' x 2) ; NB 'dashes' (') to indicate source of numbers. Write up brackets (3x 2)(x 3) -2)(x + 3) Next we notice that the '6' is negative, so the two signs are different (+/-) The '7x' is positive , so the larger number takes the positive sign . Hence (3x - 2)(x + 3)


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That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: 1 plus or minus i where i is the square root of negative one.


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There are no rational factors.


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Is y minus 3 a factor of y to the fourth power plus 2 y to the second power minus 4?

No.