Using remainder theorem , factorise completely ....
3x³ + 2x² - 19x + 6
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RUDEBOY13
Secondary SchoolMath 5+3 pts
Using remainder theorem completely factorize
3x³+2x²-19x+6
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RUDEBOY13
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3x³+2x²-19x+6
when x=2
Since the remainder is 0, x = 2 is an actual zero. This means that x - 2 is one of the factors. The numbers on the bottom represent the coefficients of the other factor. Notice that the exponents are reduced by 1.
3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3)
You can factor 3x² + 8x - 3 further.
3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3) = (x - 2)(3x - 1)(x + 3)
Set each factor to 0 and solve.
x - 2 = 0
x = 2
3x - 1 = 0
3x = 1
3x / 3 = 1 / 3
x = 1 / 3
x + 3 = 0
x = -3
ANSWER: {-3, 1 / 3, 2}
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