Question

using factor theorem show that (x-3) is a factor of x3- 7x2+15x-9 and factorise the given expression
completely

Answer 1

$\underline {\pink {Factor \: Theorem :}}$

If p(x) is a polynomial of degree n ≥ 1 and "a" is any real number, then

• (x-a) is a factor of p(x) , if p(a) = 0
• and it's converse " If (x-a) is a factor of a polynomial p(x) then p(a) = 0.

**********************

$Let \: p(x) = x^{3} - 7x^{2} + 15x - 9$

$g(x) = x - 3 ,$

$The \:zero \: of \:g(x) = 3$

$Then \: p(3) = 3^{3} - 7\times 3^{2} + 15\times 3 - 9 \\= 27 - 63 + 45 - 9 \\= 72 - 72 \\= 0$

$So, By \: Factor \: theorem , (x-3) \: is \:a \\factor \: of \:p(x)$

x-3) x³-7x²+15x-9( x²-4x+3

***** x³-3x²

______________

********* -4x²+15x

********* -4x²+12x

_______________

*************** 3x -9

*************** 3x - 9

_________________

***Remainder (0 )

$p(x) = (x-3)(x^{2}-4x+3) \\= (x-3)(x^{2} - 1x - 3x + 3)\\= (x-3)[ x(x-1)-3(x-1)] \\= (x-3)(x-1)(x-3)$

Therefore.,

$\red {x^{3} - 7x^{2} + 15x - 9}\\\green {= (x-3)(x-1)(x-3)}$

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Answer 2

Answer:

 p(x) is a polynomial of degree n ≥ 1 and "a" is any real number, then

(x-a) is a factor of p(x) , if p(a) = 0

and it's converse " If (x-a) is a factor of a polynomial p(x) then p(a) = 0.

**********************

Let \: p(x) = x^{3} - 7x^{2} + 15x - 9Letp(x)=x3−7x2+15x−9

g(x) = x - 3 ,g(x)=x−3,

The \:zero \: of \:g(x) = 3Thezeroofg(x)=3

\begin{gathered} Then \: p(3) = 3^{3} - 7\times 3^{2} + 15\times 3 - 9 \\= 27 - 63 + 45 - 9 \\= 72 - 72 \\= 0 \end{gathered}Thenp(3)=33−7×32+15×3−9=27−63+45−9=72−72=0

\begin{gathered} So, By \: Factor \: theorem , (x-3) \: is \:a \\factor \: of \:p(x) \end{gathered}So,ByFactortheorem,(x−3)isafactorofp(x)

x-3) x³-7x²+15x-9( x²-4x+3

***** x³-3x²

______________

********* -4x²+15x

********* -4x²+12x

_______________

*************** 3x -9

*************** 3x - 9

_________________

***Remainder (0 )

\begin{gathered} p(x) = (x-3)(x^{2}-4x+3) \\= (x-3)(x^{2} - 1x - 3x + 3)\\= (x-3)[ x(x-1)-3(x-1)] \\= (x-3)(x-1)(x-3) \end{gathered}p(x)=(x−3)(x2−4x+3)=(x−3)(x2−1x−3x+3)=(x−3)[x(x−1)−3(x−1)]=(x−3)(x−1)(x−3)

Therefore.,

\begin{gathered} \red {x^{3} - 7x^{2} + 15x - 9}\\\green {= (x-3)(x-1)(x-3)} \end{gathered}x3−7x2+15x−9=(x−3)(x−1)(x−3)

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