Question

show that (x-3) is a factor of the polynomial xcube
$show \: that \: {x - 3} \: is \: a \: factor \: of \: the \: polynomil \: xcube - 3 {x }^{2} + 4x - 12$
-3xsq+4x-12.​

Answer 1

Answer:

$\mathtt\green{SOLUTION:-}$

$\mathtt{x - 3 = 0}$

$\mathtt{x = 0 + 3}$

$\mathtt{x = 3}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt{NOW \: SUBSTITUTE \: THE \: VALUE \: OF \: X}$

$\mathtt \green{ = {x}^{3} - {3x}^{2} + 4x - 12 }$

$\mathtt \green{ = {3}^{3} - 3 \times {3}^{2} + 4 \times 3 - 12 }$

$\mathtt \green{ = 27 - 27 + 12 - 12}$

$\mathtt \green{ = 0}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt{YES, X-3 \: IS \: THE \: FACTOR \: OF \: THIS \: POLYNOMIAL}$

Answer 2

$\red{\large\mathfrak {Question}}$

$\sf \: Show \: that \: \: {x - 3} \: is \: a \: factor \: of \: the \: polynomial \: {x}^{3} - 3 {x }^{2} + 4x - 12$

$\green{\large\mathfrak {solution}}$

Let p(x) = x³ - 3x² + 4x - 12

and g(x) = x-3

$\sf \: (x - \alpha ) = (x - ( - 3))$

$⇒ \sf \: (x - \alpha ) = (x + 3)$

A/q remainder theorum :—

$\sf \: if \: g(x) \: is \: a \: factor \: of \: p(x) \: then \:p ( - \ alpha ) = 0$

⇒ p(3) = (3)³ - 3(3)² + 4(3) - 12

∴ p(3)=0

= 27-3×9+12-12

= 27-27+12-12

= 0

$\bold{\boxed{\blue{∵p(x) = 0 }}}$

Hence proved that g(x) is a factor of p(x).

$\pink{\large\mathfrak { \: \: \: \: !! \: hope \: it \: helps \: you \:!!}}$