remainder theorem

Without actually dividing prove that x⁴ + 2x³ – 2x² + 2x – 3 is exactly divisible by x² + 2x – 3 

Remainder theorem

Show that x² + qx + 1 and x³ + px² + qx + 1 have a common factor of the form (x + a) when

(p – 1)²- q(p – 1) + 1 = 0

remainder theorem

If x³ + px + r and 3x² + p have a common factor, 

then p³/27 + r²/4 = 0

Remainder theorem

Show that: (c + a - 2b)d²+ (a + b - 2c) d + (b + c - 2a) is exactly divisible by (d - 1).

Remainder Theorem - ICSE 2000

remainder theorem ICSE 2004

Use the factor theorem to factorize completely x³ + x² - 4x – 4. 

ICSE 2004

 

Remainder theorem - ICSE 2002

Show that (x-3) is a factor of x³ - 7x² + 15x – 9.

Hence completely factorize the expression. 

ICSE 2002

 

             

Remainder theorem

                                            

Remainder Theorem

Show that (x - 1) is a factor of x³ – 7x ² + 14x – 8. Hence completely factorize

                                                           (ICSE 2007)

Answer:

Let: F(x) = x³ – 7x ² + 14x – 8

Since (x - 1) is a factor of F(x), therefore, according to Remainder theorem,

we have, F (1) = 0

Now, F (1) = (1)³ - 7(1)² + 14(1) – 8

Or, F(1) =1 – 7 +14 - 8 = 0

So, (x – 1) is a factor.

Now we will divide (x³ – 7x ² + 14x – 8)/(x – 1)

(x - 1) // x³ – 7x ² + 14x – 8 //  x ² - 6x + 8

             x³ - x ² 

               - 6 x ² + 14x

              - 6 x ² +  6x

                            8x – 8

                           8x – 8

                           0     0  

(x ² - 6x +8) is the other factor of F(x). We will further simplify, thus:

x ² - 6x +8 = x ² - 4x - 2x + 8 = x(x – 4) – 2(x – 4) = (x – 2)(x – 4)

So, x³ – 7x ² + 14x – 8  = (x – 1 )(x – 2)(x – 4)