Question

factorise using identity 2x^3+28x^2+98x​

Answer 1

Answer:

i)

$2x(x + 7)(x + 7)$

ii)

$(a {x }^{2} + b {y}^{2} )(a {x }^{2} + b {y}^{2} )$

Step-by-step explanation:

i)

$2 {x}^{3} + 28 {x}^{2} + 98x \\ = )2x( {x}^{2} + 14x + 49) \\ = )2x(( {x)}^{2} + (2 \times 7 \times x) + ( {7)}^{2} ) \\ = )2x(x + 7 {)}^{2} \\ = )2x (x + 7)(x + 7)$

ii)

${a}^{2} {x}^{4} + 2ab {x}^{2} {y}^{2} + {b}^{2} {y}^{4} \\ = )((a {x}^{2} {)}^{2} + (2 \times a {x}^{2} + b {y}^{2} ) + (b {y}^{2} {) }^{2} \\ = )(a {x}^{2} + b {y}^{2} {)}^{2} \\ = )(a {x}^{2} + b {y} ^{2} )(a {x}^{2} + b {y} ^{2} )$

Hope it helped

Answer 2

Answer:

(i)2x^3+28x^2+98x=0

2x(x^2+14x+49)=0

Now this came as a^2+2ab+b^2

2x(x+7)(x+7)=0

(ii)By simply comparing we get (ax^2+by^2)^2=0

Step-by-step explanation: