If l and b are zeroes of the polynomial f (x) =4x power 2 -6x +2,find

1.	If l and b are zeroes of the polynomial f (x) =4x power 2 -6x +2,find the value of l3 +b3
Answer» °°QUESTION:➡If l and b are zeroes of the POLYNOMIAL f (X) =4x power 2 -6x +2,find the value of L3 +b3...☢☢ °°TO FIND:➡l^3+b^3☢☢☢☢ STEP by step explanation↪ $\bf\ \: here \: \\ \\ \bf\ \underline{ f(x) \implies {4x}^{2} - 6x + 2 = 0} \\ \\ \bf\ :\implies {4x}^{2} - (4x + 2x) + 2 = 0 \: \: \: \: \\ \\ \boxed{2 \times {4x}^{2} = {8x}^{2} = \underline{2x \: 2x} \: 2x (4x + 2x = 6x \: and \: 4x \times 2x = {8x}^{2} } \\ \\ \bf\ :\implies {4x}^{2} - 4x - 2x + 2 = 0 \\ \\ \bf\ :\implies4x(x - 1) - 2(x - 1) = 0 \\ \\ \bf\ :\implies(4x - 2)(x - 1) = 0 \\ \\ \bf\underline\red{zeros \: are } \\ \\ \bf\underline\blue{4x - 2 = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \bf\underline\green{x - 1 = 0} \\ \\ \bf\underline\blue{ \implies \: x = \frac{2}{4}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \bf\underline\green{x = 1}\\ \\ \bf\underline\blue{ \implies \: x = \frac{1}{2} } \\ \\ \bf\ if \: \bf\red{ l = \frac{1}{2}} \: \: \: and \: \: \bf\pink{b = 1} \\ \\ \bf\underline\red{then \: \: \: \: {l}^{3} = ( { \frac{1}{2} }^{3}) } \\ \\ \red{\boxed{\mapsto \frac{1}{8} }} \\ \\ \bf\underline\blue{ {b}^{3} = {1}^{3} } \\ \\ \blue{ \boxed{ \mapsto1}} \\ \\ \bf\red{l^3+b^3=\frac{1}{8}+1} \\ \\ \bf\red{ \implies \frac{1+9}{8}} \\ \\ \bf\red{ \implies \frac{10}{8}}\\ \\ \bf\red{ \implies \frac{5}{4}} \\ \\ \large\boxed{ \fcolorbox{lime}{yellow}{hope \: it \: help \: you}}$

If l and b are zeroes of the polynomial f (x) =4x power 2 -6x +2,find the value of l3 +b3

Answer»

°°QUESTION:➡If l and b are zeroes of the POLYNOMIAL f (X) =4x power 2 -6x +2,find the value of L3 +b3...☢☢

°°TO FIND:➡l^3+b^3☢☢☢☢

STEP by step explanation↪

$\bf\ \: here \: \\ \\ \bf\ \underline{ f(x) \implies {4x}^{2} - 6x + 2 = 0} \\ \\ \bf\ :\implies {4x}^{2} - (4x + 2x) + 2 = 0 \: \: \: \: \\ \\ \boxed{2 \times {4x}^{2} = {8x}^{2} = \underline{2x \: 2x} \: 2x (4x + 2x = 6x \: and \: 4x \times 2x = {8x}^{2} } \\ \\ \bf\ :\implies {4x}^{2} - 4x - 2x + 2 = 0 \\ \\ \bf\ :\implies4x(x - 1) - 2(x - 1) = 0 \\ \\ \bf\ :\implies(4x - 2)(x - 1) = 0 \\ \\ \bf\underline\red{zeros \: are } \\ \\ \bf\underline\blue{4x - 2 = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \bf\underline\green{x - 1 = 0} \\ \\ \bf\underline\blue{ \implies \: x = \frac{2}{4}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \bf\underline\green{x = 1}\\ \\ \bf\underline\blue{ \implies \: x = \frac{1}{2} } \\ \\ \bf\ if \: \bf\red{ l = \frac{1}{2}} \: \: \: and \: \: \bf\pink{b = 1} \\ \\ \bf\underline\red{then \: \: \: \: {l}^{3} = ( { \frac{1}{2} }^{3}) } \\ \\ \red{\boxed{\mapsto \frac{1}{8} }} \\ \\ \bf\underline\blue{ {b}^{3} = {1}^{3} } \\ \\ \blue{ \boxed{ \mapsto1}} \\ \\ \bf\red{l^3+b^3=\frac{1}{8}+1} \\ \\ \bf\red{ \implies \frac{1+9}{8}} \\ \\ \bf\red{ \implies \frac{10}{8}}\\ \\ \bf\red{ \implies \frac{5}{4}} \\ \\ \large\boxed{ \fcolorbox{lime}{yellow}{hope \: it \: help \: you}}$