Question #6a4c5

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Question #6a4c5

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Answer 1 · 2017-02-04T22:19:26

$y^{2} - 22 y + 121 = (y - 11) (y - 11)$ $y^2 - 22y + 121 = (y - 11)(y- 11)$

Explanation:

$y^{2} - 22 y + 121$ $y^2 -22y+121$

By splitting the middle term i.e. -22 in such a way so, that the sum of those two numbers would be -22 and their product would be 121. Therefore, these two numbers are -11 and -11 . Here, both conditions are applied.

$y^{2} - 11 y - 11 y + 121$ $y^2 - 11y - 11y + 121$

$y (y - 11) - 11 (y - 11)$ $y(y - 11) - 11(y -11)$

$(y - 11) (y - 11)$ $(y - 11)(y- 11)$

Alternative Method:

$y^{2} - 22 y + 121$ $y^2 - 22y + 121$

By using the identity -

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

Here , $a = y, b = 11$ $a = y , b = 11$

${(y)}^{2} - 2 (y) (11) + {(11)}^{2}$ $(y)^2 - 2(y)(11) + (11)^2$

$(y - 11) (y - 11)$ $(y - 11)(y -11)$

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Question #6a4c5

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