Question

p(x)=x²+x2-10x+8 (b) Using long division express p(x) as product of linear factor (e) Hence solve x¹+x²-10x +8=0 Qu. 3. The polynomial p/x) is given (a) Show that x-2 is a factor of p(x)​

Answer 1

b. To express p(x) as a product of linear factors using long division, we divide p(x) by (x - 2) and get a quotient of (x + 5) and a remainder of 0. This means that p(x) = (x - 2)(x + 5)

e. To solve p(x) = 0, we set each factor equal to zero and find the solutions:

x - 2 = 0, x = 2

x + 5 = 0, x = -5

Thus, the solutions to the equation p(x) = 0 are x = 2 and x = -5.

a. To show that x-2 is a factor of p(x), we can use synthetic division or polynomial long division.

let's use polynomial long division to divide p(x) by (x-2)

p(x) = x²+x²-10x+8

| x-2 |

x²-2x+x

| -x²-x²+10x+8

--------------

0

As we can see, after dividing p(x) by (x-2) we get remainder as 0, which implies that (x-2) is a factor of p(x)

Hence, we can say that x-2 is a factor of p(x)