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Division of a Polynomial by a Monomial

a2-b2=a+ba-b ...

Question

$a^{2} - b^{2} = (a + b) (a - b)$ then factor of $1 + 2 a b - (a^{2} + b^{2})$ are

[1 + a + b] [1 - a - b]

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[1 + a - b] [1 - a + b]

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[1 + a + b] [1 + a - b]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[1 + a + b] [1 + a + b]

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Solution

The correct option is
A
$[1 + a + b] [1 - a - b]$

$\Rightarrow y = 1 + 2 a b - (a^{2} + b^{2})$

$y = 1 - [a^{2} + b^{2} - 2 a b]$
$y = 1 - (a - b)^{2}$
$y = (1)^{2} - (a - b)^{2}$
Given $a^{2} - b^{2} = (a - b) (a + b)$
then
$y = (1 - a + b) (a + a - b)$

Suggest Corrections

Similar questions

a^{2} - b^{2} = (a + b) (a - b)

1 + 2 a b - (a^{2} + b^{2})

A = (\begin{matrix} 1 & - 1 2 & 1 \end{matrix}), B = (\begin{matrix} a & 1 b & - 1 \end{matrix})

(A + B)^{2} = A^{2} + B^{2} + 2 A B

a = b = 1

a = b

a^{2} = a b

a^{2} - b^{2} = a b - b^{2}

(a + b) (a - b) = b (a - b)

\frac{b (a - b)}{a - b}

a + b = b

1 + 1 = 1

2 = 1

A = [\begin{matrix} 2 & - 1 0 & 1 \end{matrix}]

B = [\begin{matrix} 1 & 0 - 1 & - 1 \end{matrix}],

(A + B)^{2} = A^{2} + A B + B A + B^{2} = A^{2} + 2 A B + B^{2}

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

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