Question

Divide x^4-7x^3+17x^2-17x+6 by x^2-4x-3 and find its factors​

Answer 1

... + 1 Ans. No 3. Obtain all zeroes of f(x) = x4–3x3–x2 + 9x – 6 if two of its zeroes are ... Ans. x2–4x – 32 9. ... Ans. k = –9, factors (x2 + 5x + 6); –2, –3 4 ...

pls mark Brainliest

Answer 2

Answer:

since 1and 2 are the zero of the polynomial p(x)=x4-7x3+17x2-17x+6,

therefore, x-1 and x-2 are factor of p(x) {by factor theorem}

now,(x-1) (x-2) =x4 -7x3+17x2-17x+6

x2-3x+2 =x4 -7x3+17x2-17x+6

dividing p(x) by x2-3x+2,

we get quotient as x2-4x +3=(x-1) (x-3) [by factorising x2 -4x+3]

p(x)=(x-1) (x-3) (x-1) (x-2)

for the zeroes of p(x),equat p(x) to zero

p(x)=0,

(x-1) (x-3) (x-1) (x-2) =0

x =1,1,2and3

hence all four zeroes of p(x) are 1,1,3and 2