Question

by long division method show that (x-3)factor of 2x^4+3x^3-26x2-5x+6

Answer 1

Solution :

p(x) = 2x⁴ + 3x³ - 26x² - 5x + 6

g(x) = x - 3

After dividing we obtained :

• Quotient = 2x³ + 9x² + x - 2

• Remainder = 0

Refer to the attachment!

Verification :

★ Dividend = Divisor × Quotient + Remainder

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = (x - 3) × 2x³ + 9x² + x - 2 + 0

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = x(2x³ + 9x² + x - 2) - 3(2x³ + 9x² + x - 2) + 0

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = (2x⁴ + 9x³ + x² - 2x) - (6x³ + 27x² + 3x - 6) + 0

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = 2x⁴ + 9x³ + x² - 2x - 6x³ - 27x² - 3x + 6

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = 2x⁴ + 9x³ - 6x³ - 27x² + x² - 2x - 3x + 6

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = 2x⁴ + 3x³ - 26x² - 5x + 6

$\therefore$ (x-3) is factor of the polynomial 2x⁴ + 3x³ - 26x² - 5x + 6

Answer 2

Answer:

❖ $\fbox\purple{\sf Quotient = 2x^3 + 9x^2 + x - 2}$

❖ $\fbox\purple{\sf Remainder = 0}$

Step-by-step explanation:

Let's verify it,

❖ Dividend = Divisor × Quotient + Remainder

• Divided = 2x⁴ + 3x³ - 26x² - 5x + 6
• Divisor = (x - 3)
• Quotient = 2x³ + 9x² + x - 2
• Remainder = 0

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = (x - 3)(2x³ + 9x² + x - 2) + 0

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = 2x⁴ + 9x³ + x² - 2x - 6x³ - 27x² - 3x + 6

→ 2x⁴ + 3x³ - 26x² - 5x + 6 = 2x⁴ + 3x³ - 26x² - 5x + 6

$\fbox{\sf Hence, Verified. }$

$\fbox{\sf \therefore (x+3)\:is\: a\: factor\: of\: polynomial\:2x^4 + 3x^3 - 26x^2 - 5x + 6 (as\:given\:in\:the\:question)}$

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