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

### 2 thoughts on “show that (x-3) is a factor of the polynomial xcube<br />$$show \: that \: {x – 3} \: is \: a \: factor \: of \: the \: polynom” 1. [tex]\red{\large\mathfrak {Question}}$$

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

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

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).

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

$$\mathtt\green{SOLUTION:-}$$

$$\mathtt{x – 3 = 0}$$

$$\mathtt{x = 0 + 3}$$

$$\mathtt{x = 3}$$

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

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

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

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

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

$$\mathtt \green{ = 0}$$

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

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