Find the quotient following the steps. write your answer in your mathematics notebook.
Answer:
- x–4
- [tex] \frac{6x^4y^7–4x^2y^5+3}{y^3} [/tex]
- y–5
- [tex]x = \frac{ - {a}^{3} + 4 {a}^{2} + 3 }{2} [/tex]
Step-by-step explanation:
1.
[tex] \frac{ {x}^{3} - 4 {x}^{2} }{ {x}^{2} } = x - 4 \\ \\ {x}^{3} - 4 {x}^{2} \\ {x}^{3} = x {x}^{2} \\ x {x}^{2} - 4 {x}^{2} \\ {x}^{2} (x - 4) \\ \frac{{x}^{2} (x - 4)}{ {x}^{2} } \\ \frac{ \cancel{{x}^{2}} (x - 4)}{ \cancel{{x}^{2} }} \\ \underline \green{x - 4}[/tex]
2.
[tex] \frac{18 {x}^{5} {y}^{7} - 12 {x}^{3} {y}^{5} + 9x}{3x {y}^{3} } = \frac{ 6{x}^{4} {y}^{7} - 4 {x}^{2} {y}^{5} + 3}{{y}^{3} } \\ \\ 18 {x}^{5} {y}^{7} - 12 {x}^{3} {y}^{5} + 9x \\ {x}^{5} {y}^{7} = x {x}^{4} - 12x {x}^{2} + 9x \\ 6 \times 3x {x}^{4} + 4 \times 3x {x}^{2} + 3 \times 3x \\ \frac{3x(6 {x}^{4} {y}^{7} - 4 {x}^{2} {y}^{5} + 3)}{3x {y}^{3} } \\ \frac{ \cancel{3x}(6 {x}^{4} {y}^{7} - 4 {x}^{2} {y}^{5} + 3)}{ \cancel{3x} {y}^{3} } \\ \underline \green{ \frac{ 6 {x}^{4} {y}^{7} - 4 {x}^{2} {y}^{5} + 3}{ {y}^{3} } }[/tex]
3.
[tex] \frac{ {y}^{2} - 3y - 10}{y + 5} = y - 5 \\ \\ {y}^{2} - 3y - 10 \\ ( {y}^{2} + 2y) + ( - 5y - 10) \\ {y}^{2} + 2y = yy + 2y \\ y(y + 2) \\ - 5y - 10 = - 5y - 5 \times 2 \\ - 5(y + 2) \\ y(y + 2) - 5(y + 2) \\ (y + 2)(y - 5) \\ y(y + 2) - 5(y + 2) \\ \frac{\cancel{ (y + 2)}(y - 5) }{ \cancel{y + 5}} \\ \underline \green{y - 5}[/tex]
4.
[tex] \frac{ {a}^{3} - 4 {x}^{2} + 2x - 3 }{a + 2 } = \:x = \frac{ - {a}^{3} + 4 {a}^{2} + 3 }{2} \\ \\ \frac{ {a}^{3} - 4 {x}^{2} + 2x - 3 }{a + 2 } = 0 \: there \: roots \: are \: intercepts \: with \: the \: x - axis \: (y = 0) \\ {a}^{3} - 4 {x}^{2} + 2x - 3 = 0 \\ {a}^{3} - 4 {a}^{2} + 2x - 3 - ( {a}^{3} - 4 {a}^{2} ) = 0 - ({a}^{3} - 4 {a}^{2} ) \\ 2x - 3 = - ( {a}^{3} - 4 {a}^{2} ) \\ 2x - 3 + 3 = - ({a}^{3} - 4 {a}^{2} + 3 )\\ 2x = - {a}^{3} + 4 {a}^{2} + 3 \\ 2x/2 = - {a}^{3} /2 + 4 {a}^{2} /2 + 3/2 \\ \underline \green{x = \frac{ - {a}^{3} + 4 {a}^{2} + 3 }{2} }[/tex]
[tex] \\ [/tex]
Hope it's help
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