Question

Can someone solve this, with explanation please

Answer 1

Ques 1. Sol Upper

Ques 2.

Sol. Consider the equation

x3 – 3x2 – 9x – 5…………………..(1)

The equation can be simplified as

x3 + x2 – 4x2 – 4x – 5x – 5

x2 (x + 1) – 4x (x + 1) -5(x + 1)

(x + 1)(x2 – 4x – 5)

(x + 1)(x2 – 5x + x – 5)

(x + 1) (x – 5)(x + 1)

∴ (x + 1), (x – 5) and (x + 1) are factors of given polynomial.

Ques 3.

Sol. Let p(x)=ax

3

+bx

2

+x−6

Since (x+2) is a factor of p(x), then by Factor theorem p(−2)=0

⇒a(−2) 3

+b(−2) 2

+(−2)−6=0

⇒−8a+4b−8=0

⇒−2a+b=2

Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem p(2)=4

⇒a(2) 3 +b(2)

2+2−6=4

⇒8a+4b+2−6=4

⇒8a+4b=8

⇒2a+b=2

Adding equation (i) and (ii), we get

(−2a+b)+(2a+b)=2+2

⇒2b=4⇒b=2

Putting b=2 in (i), we get

−2a+2=2

⇒−2a=0⇒a=0

Hence, a=0 and b=2