show that (x-3) is a factor of the polynomial xcube
[tex]show \: that \: {x – 3} \: is \: a \: factor \: of \: the \: polynom

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

2 thoughts on “show that (x-3) is a factor of the polynomial xcube<br />[tex]show \: that \: {x – 3} \: is \: a \: factor \: of \: the \: polynom”

  1. [tex]\red{\large\mathfrak {Question}}[/tex]

    [tex] \sf \: Show \: that \: \: {x – 3} \: is \: a \: factor \: of \: the \: polynomial \: {x}^{3} – 3 {x }^{2} + 4x – 12[/tex]

    [tex]\green{\large\mathfrak {solution}}[/tex]

    Let p(x) = x³ – 3x² + 4x – 12

    and g(x) = x-3

    [tex] \sf \: (x – \alpha ) = (x – ( – 3))[/tex]

    [tex]⇒ \sf \: (x – \alpha ) = (x + 3)[/tex]

    A/q remainder theorum :—

    [tex] \sf \: if \: g(x) \: is \: a \: factor \: of \: p(x) \: then \:p ( – \ alpha ) = 0[/tex]

    ⇒ p(3) = (3)³ – 3(3)² + 4(3) – 12

    ∴ p(3)=0

    = 27-3×9+12-12

    = 27-27+12-12

    = 0

    [tex]\bold{\boxed{\blue{∵p(x) = 0 }}}[/tex]

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

    [tex]\pink{\large\mathfrak { \: \: \: \: !! \: hope \: it \: helps \: you \:!!}}[/tex]

  2. Answer:

    [tex]\mathtt\green{SOLUTION:-}[/tex]

    [tex] \mathtt{x – 3 = 0}[/tex]

    [tex]\mathtt{x = 0 + 3}[/tex]

    [tex] \mathtt{x = 3}[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

    [tex]\mathtt{NOW \: SUBSTITUTE \: THE \: VALUE \: OF \: X}[/tex]

    [tex] \mathtt \green{ = {x}^{3} – {3x}^{2} + 4x – 12 }[/tex]

    [tex] \mathtt \green{ = {3}^{3} – 3 \times {3}^{2} + 4 \times 3 – 12 }[/tex]

    [tex] \mathtt \green{ = 27 – 27 + 12 – 12}[/tex]

    [tex] \mathtt \green{ = 0}[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

    [tex]\mathtt{YES, X-3 \: IS \: THE \: FACTOR \: OF \: THIS \: POLYNOMIAL}[/tex]

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