(2x³-5x²-x+7) by (3-2x+4x²)
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Step-by-step explanation:
x³ + 5x² - 5x - 2
= 2x³ - 2x² + 7x² - 7x + 2x - 2
= 2x²(x - 1) + 7x(x - 1) + 2(x - 1)
= (2x² + 7x + 2)(x - 1)
Here we have to factors (2x² + 7x + 2) and (x - 1)
one zero of given Polynomial , x - 1 = 0 ⇒ x = 1 ( integer )
other two zeros of given polynomial , 2x² + 7x + 2 = 0 ⇒ x = (-7 ± √33)/4 [ irrational numbers ]
Hence, internal zero of polynomial has only one e.g., x = 1
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