Question

49x^4 + 168x^2y^2 + 144y^4

ans is (7x2 +12y2)2
​

Answer 1

Step-by-step explanation:

Given that:

$49 {x}^{4} + 168 {x}^{2} {y}^{2} + 144 {y}^{4}$

To find: Factors

Solution: To find the factors of given polynomial

relate the term with the identity

$\bold{( {a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}}$

Now convert given polynomial

$(7 {x}^{2})^{2} + 2 \times 7 \times 12{x}^{2} {y}^{2} + ( {12}{y}^{2} )^{2} \\ \\ here \: one \: can \: compare \: with \: identity \\ \\ a = (7 {x}^{2}) \\ \\ b = ( {12}{y}^{2} )$

Thus,

$( {7x}^{2} + 12 {y}^{2} )^{2}$

So,

$\bold{49 {x}^{4} + 168 {x}^{2} {y}^{2} + 144 {y}^{4} = ( {7x}^{2} + 12 {y}^{2} )^{2}}$

Hope it helps you.

Answer 2

GIVEN :

The polynomial is [tex]49x^4 + 168x^2y^2 + 144y^4 [/tex]

TO FIND :

The factors of the given polynomial.

SOLUTION :

Given polynomial is [tex]49x^4 + 168x^2y^2 + 144y^4 [/tex]

By Factorisation we can find the factors as below :

Now solving given polynomial,

[tex]49x^4 + 168x^2y^2 + 144y^4 [/tex]

$=7^2x^4+168x^2y^2+12^2y^4$

By using the exponent power rule :

$(a^m)^n=a^{mn}$

$=7^2(x^2)^2+168x^2y^2+12^2(y^2)^2$

By using the exponent power rule :

$a^mb^m=(ab)^{m}$

$=(7x^2)^2+2(7x^2)(12y^2)+(12y^2)^2$

By using the Algebraic identity :

$(a+b)^2=a^2+2ab+b^2$

$=(7x^2+12y^2)^2$

or  $=(7x^2+12y^2)(7x^2+12y^2)$

$49x^4 + 168x^2y^2 + 144y^4=(7x^2+12y^2)^2$

∴ The factors of the given polynomial $49x^4 + 168x^2y^2 + 144y^4$ is $(7x^2+12y^2)^2$