Question

Please do the test correctly and fast please i need it urvently please...​

Answer 1

Step-by-step explanation:

1)

For 0 of (x-1)

=x-1=0

=x=1

Therefore,

$p(1) = {1}^{2} - 2(1) + 1 \: \: \: \: \: \: \: \: (by \: remainder \: theorem) \\ = > 1 - 2 + 1 \\ = > 0$

2)

$(x - 1) - ( {x}^{2} - 1) \\ = > x - 1 - {x}^{2} - 1 \\ = > - {x}^{2} + x - 2$

Therefore, the coefficient of x² is -1

3)

$(x - \frac{1}{x} )(x + \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } ) \: \: \: \: \: \: \: \\ using \: identity \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = > ( {x}^{2} - \frac{1}{ {x}^{2} })( {x}^{2} + \frac{1}{ {x}^{2} } ) \\ \\ = > ( {x}^{4} - \frac{1}{ {x}^{4} })</p><p>$

4)

$p(x) = {x}^{19} + 1 \\ g(x) = x + 1$

For zero of g(x)

x+1=0

=x=-1

Therefore,

$p( - 1) = {( - 1)}^{19} + 1 \: \: \: \: \: (by \: remainder \: theorem) \\ = > - 1 + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: </p><p>$

Since g(x) is exactly divisible by p(x)

Therefore g(x) is a factor of p(x)

[By factor theorem]

5)

a)

b)

$125 {n}^{3} + 27 {y}^{3} \\ = > {(5n)}^{3} + {(3y)}^{2} \\ = > (5n + 3y)( {(5n)}^{2} - (5n)(3y) + {(3y)}^{2} ) \\ = > (5n + 3y)(25 {n}^{2} - 15ny + 9 {y}^{2} ) \\ </p><p></p><p>$

c)

${(x - y)}^{3} + {(y - z)}^{3} + (z - x)^{3} \\ = > since(x - y) + (y - z) + (z - x) = 0 \\ therefore \: {(x - y)}^{3} + {(y - z)}^{3} + (z - x)^{3} = 3(x - y)(y - z)(z - x) \\ = > 3(xy - xz - {y}^{2} + yz)(z - x) \\ = > (3)(xyz + {x}^{2}y - x {z}^{2} + {x}^{2} z - {y}^{2} z + {y}^{2} x + y {z}^{2} - xyz) \\ = > 3( - {x}^{2}y + {x}^{2}z - x {z}^{2} - {y}^{2}z + {y}^{2} x + y {z}^{2} ) \\ = > 3( - {x}^{2} y + {x}^{2} z - {y}^{2} x + {y}^{2} z - x {z}^{2} + y {z}^{2} ) \\ = > 3( - {x}^{2} (y - z) - {y}^{2} (x - z) - {z}^{2} (x - y)) \\ = > 3( - {x}^{2} - {y}^{2} - {z}^{2} )(y - z)(x - z)(x - y)$

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