Question

if x^2-3x+2is factor of x^4-ax^2+b find the value of a,b​

Answer 1

$\large{ \bold{p(x) = {x}^{2} - 3x + 2}} \\ \\ \large{ \bold{p(x) = {x}^{2} - x - 2x + 2 }}\\ \\ \large{ \bold{p(x) = x(x - 1) - 2(x - 1) }}\\ \\ \large{ \bold{p(x) = (x - 1)(x - 2)}} \\ \\ \large{ \bold{also}} \\ \\ \large{ \bold{f(x) = {x}^{4} - a {x}^{2} + b}} \\ \\ \large{ \bold{given}} \\ \\ \large{ \bold{p(x) \: is \: a \: factor \: of \: p(x)}} \\ \\ = > \large{ \bold{f(1) = 0 \: \: \: and \: \: \: f(2) = 0}} \\ \\ = > \large{ \bold{f(1) = {1}^{4} - (a ){1}^{2} + b}} \\ \\ = > \large{ \bold{1 - a + b = 0}} \\ \\ = > \large{ \bold{a - b = 1}} \\ \\ \large{ \bold{and}} \\ \\ = > \large{ \bold{f(2) = {2}^{4} - (a) {2}^{2} + b}} \\ \\ = > \large{ \bold{16 - 4a + b = 0}} \\ \\ = > \large{ \bold{4a - b = - 16}} \\ \\ = > \large{ \bold{3a + a - b = - 16}} \\ \\ = > \large{ \bold{3a = - 17}} \\ \\ = > \large{ \bold{a = - \frac{17}{3} }} \\ \\ = > \large{ \bold{b = - \frac{20}{3} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \color{red}{\boxed{\large{ \bold{hope \: it \: helps}}}}$

Answer 2

$\large{ \bold{p(x) = {x}^{2} - 3x + 2}} \\ \\ \large{ \bold{p(x) = {x}^{2} - x - 2x + 2 }}\\ \\ \large{ \bold{p(x) = x(x - 1) - 2(x - 1) }}\\ \\ \large{ \bold{p(x) = (x - 1)(x - 2)}} \\ \\ \large{ \bold{also}} \\ \\ \large{ \bold{f(x) = {x}^{4} - a {x}^{2} + b}} \\ \\ \large{ \bold{given}} \\ \\ \large{ \bold{p(x) \: is \: a \: factor \: of \: p(x)}} \\ \\ = > \large{ \bold{f(1) = 0 \: \: \: and \: \: \: f(2) = 0}} \\ \\ = > \large{ \bold{f(1) = {1}^{4} - (a ){1}^{2} + b}} \\ \\ = > \large{ \bold{1 - a + b = 0}} \\ \\ = > \large{ \bold{a - b = 1}} \\ \\ \large{ \bold{and}} \\ \\ = > \large{ \bold{f(2) = {2}^{4} - (a) {2}^{2} + b}} \\ \\ = > \large{ \bold{16 - 4a + b = 0}} \\ \\ = > \large{ \bold{4a - b = - 16}} \\ \\ = > \large{ \bold{3a + a - b = - 16}} \\ \\ = > \large{ \bold{3a = - 17}} \\ \\ = > \large{ \bold{a = - \frac{17}{3} }} \\ \\ = > \large{ \bold{b = - \frac{20}{3} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \color{blue}{\boxed{\large{ \bold{hope \: it \: helps}}}}$