Question

Multiply (2x - 5x² - x + 7) by (3 - 2x + 4x²)​

Answer 1

Answer:

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

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

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

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

Step-by-step explanation: