Question

divide the polynomial by using factor method (y²-y-42) by (y-7).
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. please give full solution.​

Answer 1

Answer:

$\sf \dfrac{y^2 - y - 42}{y-7} = y+6$

Step-by-step explanation:

Divide the polynomial (y² - y - 42) by (y - 7).

By using factorisation method,

$\implies \sf \dfrac{y^2 - y - 42}{y - 7}$

$\implies \sf \dfrac{y^2 - 7y + 6y - 42}{y - 7}$

$\implies \sf \dfrac{y(7-y)+6(y-7)}{y-7}$

$\implies \sf \dfrac{(y-7)(y+6)}{(y-7)}$

Cancelling the the numerator by denominator with same value (y - 7).

$\large{\underline{\boxed{\sf \dfrac{y^2 - y - 42}{y-7} = y+6}}}$

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Know more :-

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$