Question

Factorise: x3 – 10x2 – x +10.

Answer 1

Answer:

⇒x³-10x²-x+10

⇒x²(x-10)-1(x-10)

⇒x²-1(x-10)

⇒(x+1)(x-1)(x-10)

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Answer 2

Answer:

x³ - 10x² - x + 10 = (x - 1) (x - 10) (x + 1)

Step-by-step explanation:

Given cubic polynomial :

x³ - 10x² - x + 10

Method - 1 :

x³ - 10x² - x + 10

x² (x - 10) - 1(x - 10)

(x - 10) (x² - 1)

(x - 10) (x² - 1²)  

(x - 10) (x - 1) (x + 1)   [ ∵ a² - b² = (a - b) (a + b) ]

Method - 2 :

Note : If the sum of all the coefficients of the cubic polynomial is equal to zero, then (x - 1) is a factor of the given cubic polynomial.

Let's see if (x - 1) is a factor of the given cubic polynomial.

⇒ Sum of the coefficients

 = 1 + (-10) + (-1) + 10

 = 1 - 10 - 1 + 10

 = 0

The result is zero. Hence (x - 1) is a factor of the given cubic polynomial.

i.e., 1 is a zero of the given polynomial.

Now, divide the given polynomial x³ - 10x² - x + 10 by (x - 1)

        $\Large \begin{array}{c|c|c} \sf x-1 & \sf x^3-10x^2-x+10 &\sf x^2-9x-10 \\ & \sf x^3-x^2 \qquad \qquad \quad & \\ \cline{2-2} & \sf -9x^2 -x+10 & \\ & \sf -9x^2+9x \qquad & \\ \cline{2-2} & \sf \qquad \ -10x+10 & \\ & \sf \qquad \ -10x+10 \\ \cline{2-2} & \qquad \quad \sf 0 \end{array}$

Quotient = x² - 9x - 10

Now, we have to factorize x² - 9x - 10

x² - 9x - 10

x² - 10x + x - 10

x(x - 10) + 1(x - 10)

(x - 10) (x + 1)

x² - 9x - 10 = (x - 10) (x + 1)

Therefore, x³ - 10x² - x + 10 = (x - 1) (x - 10) (x + 1)