Question

$\sf \huge \green{..question..}$
Pleas refer to the image attached. Q. 12.

Spams not needed. Reported on the spot.

Quality Answers needed.

Good explanation needed.

Class 10th. ML AGARWAL. ​

Answer 1

Questions:-

(i) Show that (x - 1) is a factor of x³ - 5x² - x + 5. Hence, factorise x³ - 5x³ - x + 5.

(ii) Show that (x - 3) is a factor of x³ - 7x² + 15x - 9. Hence, factorise x³ - 7x² + 15x - 9.

Required Answers:-

(a) Given:-

• g(x) = x - 1
• p(x) = x³ - 5x² - x + 5

Solution:-

For showing:-

We have,

g(x) = x - 1

=> g(x) = 0

=> x - 1 = 0

=> x = 1

Now Putting the value of x in p(x)

p(x) = x³ - 5x² - x + 5

=> p(1) = (1)³ - 5(1)² - 1 + 5

=> p(1) = 1 - 5 - 1 + 5

=> p(1) = 1 - 1 + 5 - 5

=> p(1) = 0

Hence (x - 1) is the factor of x³ - 5x² - x + 5 [Shown]

For Factorisation:-

x³ - 5x² - x + 5

= x²(x - 5) -1(x - 5)

= (x - 5)(x² - 1)

= (x - 5)(x + 1)(x - 1)

Hence Factorised!!!!

______________________________________

(b) Given:-

• g(x) = x - 3
• p(x) = x³ - 7x² + 15x - 9

Solution:-

For showing:-

We have:-

g(x) = x - 3

=> g(x) = 0

=> x - 3 = 0

=> x = 3

Putting the value of x in p(x)

p(x) = x³ - 7x² + 15x - 9

=> p(3) = (3)³ - 7(3)² + 15(3) - 9

=> p(3) = 27 - 7(9) + 45 - 9

=> p(3) = 27 + 45 - 63 - 9

=> p(3) = 72 - 72

=> p(3) = 0

Hence (x - 3) is the factor of x³ - 7x² + 15x - 9 [Shown]

For Factorization:-

Let us divide the p(x) by g(x)

= $\sf{\dfrac{x^3 - 7x^2 + 15x - 9}{x - 3}}$

= $\begin{array}{clc} \sf{x - 3)} & \sf{x^3 - 7x^2 + 15x - 9} & \sf{(x^2 - 4x + 3} \\ & \sf{x^3 - 3x^2} & \\ & \sf{(-)} \:\:\sf{(+)}&\\ &\dfrac{\qquad\qquad\qquad\qquad\qquad}{}& \\ & \sf{\:\:\:\:\:\:\:\:-4x^2 + 15x - 9} & \\ & \sf{\:\:\:\:\:\:\:\:-4x^2 + 12x}& \\ & \sf{\:\:\:\:\:\:\:\:(+) \:\:\:\:(-)} & \\ & \dfrac{\qquad\qquad\qquad\qquad\qquad}{} & \\ &\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:3x - 9} & \\ & \sf{ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: 3x - 9}& \\ & \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(-)\:\:(+)} & \\ & \dfrac{\qquad\qquad\qquad\qquad\qquad}{}& \\ & \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0}&\end{array}$

So on dividing p(x) by g(x) we get quotient as x² - 4x + 3

Hence,

x³ - 7x² + 15x - 9 = (x - 3)(x² - 4x + 3)

On further factorising:-

(x - 3)(x² - 4x + 3)

Middle term splitting the second bracket.

= (x - 3)(x² - 3x - x + 3)

= (x - 3)[x(x - 3) -1(x - 3)]

= (x - 3)[(x - 3)(x - 1)]

= (x - 3)(x - 3)(x - 1)

= (x - 3)²(x - 1)

Hence Factorized!!!

________________________________

Answer 2

Answer:

kis ki lovedeep ki Huh... bhaiya mujhe bht rona aa rha h divya didi kinna hurt hui h or ld ki wjh se brainly ch.od rhi h means hme c.hod rhi h