Question

Factorise : x^3+7x^2-x-7​

Answer 1

Answer:

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Answer 2

Answer:

x³ + 7x² - x - 7 = (x - 1) (x + 1) (x + 7)

Step-by-step explanation:

Given cubic polynomial : x³ + 7x² - x - 7

Method - 1 :

x³ + 7x² - x - 7

x³ - x + 7x² - 7

x(x² - 1) + 7(x² - 1)

(x² - 1) (x + 7)

(x² - 1²) (x + 7)

(x - 1) (x + 1) (x + 7)  [ ∵a² - b² = (a - b)(a + b) ]

Method - 2 :

Trial and error method : Put x = 1,

 x³ + 7x² - x - 7

= (1)³ + 7(1)² - (1) - 7

= 1 + 7(1) - 1 - 7

= 1 + 7 - 1 - 7

= 0

Since, the result is zero.

1 is a zero of the given polynomial.

Hence, (x - 1) is a factor.

Divide x³ + 7x² - x - 7 by (x - 1)

    $\Large \begin{array}{c|c|c}\sf x-1&\sf \quad x^3+7x^2-x-7 \qquad &\sf x^2+8x+7\\&\sf x^3-x^2 \qquad \qquad \quad \\ \cline{2-2} & \sf \qquad \ \ 8x^2-x-7 \qquad & \\ & \sf 8x^2-8x \ \\ \cline{2-2} & \sf \qquad \qquad \quad 7x-7 & \\ & \sf \qquad \qquad \quad 7x-7 \\ \cline{2-2} & \sf \qquad \qquad \quad \: 0 \end{array}$

Remainder = 0

Quotient = x² + 8x + 7

Now, factorize the quadratic polynomial x² + 8x + 7

x² + 8x + 7

x² + x + 7x + 7

x(x + 1) + 7(x + 1)

(x + 1) (x + 7)

Therefore, (x + 1) and (x + 7) are factors of the quadratic polynomial x² + 8x + 7

⇒ x³ + 7x² - x - 7

  = (x - 1) (x² + 8x + 7)

  = (x - 1) (x + 1) (x + 7)