Without actual division, prove that ( x – 2 ) is a factor of the polynomial
3x3 – 13x2 + 8x + 12. Also, factorise it completely.
Answers
Answer:
WithoutActualdivisionproving:
\begin{gathered}Let \: p(x) = 3x^{3} - 13x^{2} + 8x + 12\:and \\g(x) = x -2\end{gathered}
Letp(x)=3x
3
−13x
2
+8x+12and
g(x)=x−2
The \: zero \:of \:g(x) \:is \: 2Thezeroofg(x)is2
\begin{gathered}Then \:p(-2) = 3(2)^{3} - 13(2)^{2} + 8 \times 2 + 12 \\= 3 \times (8) - 13 \times 4 + 16 + 12 \\= 24 - 52 + 16 + 12 \\= 52 - 52 \\= 0\end{gathered}
Thenp(−2)=3(2)
3
−13(2)
2
+8×2+12
=3×(8)−13×4+16+12
=24−52+16+12
=52−52
=0
\begin{gathered}So, \pink { by \:the \:Factor \: theorem }, \: (x-2) \:is \:a \\factor \: of \:3x^{3} - 13x^{2} + 8x + 12\end{gathered}
So,bytheFactortheorem,(x−2)isa
factorof3x
3
−13x
2
+8x+12
\underline {\blue {Using \: division\: method: }}
Usingdivisionmethod:
x-2|3x³-13x²+8x+12|3x²-7x-6
***** 3x^{3}-6x^{2}
___________________
************ - 7x²+ 8x
************ -7x²+ 14x
____________________
********************** - 6x + 12
********************** -6x + 12
______________________
************* Remainder ( 0)
\begin{gathered}Now, \: \blue { ( By\: division \: algorithm : )}\\ 3x^{3} - 13x^{2} + 8x + 12\\ = (x-2)(3x^{2}-7x-6)\\= (x-2)( 3x^{2} -9x + 2x - 6 ) \\= (x-2)[3x(x-3) + 2(x-3) ]\\= (x-2)(x-3)(3x+2)\end{gathered}
Now,(Bydivisionalgorithm:)
3x
3
−13x
2
+8x+12
=(x−2)(3x
2
−7x−6)
=(x−2)(3x
2
−9x+2x−6)
=(x−2)[3x(x−3)+2(x−3)]
=(x−2)(x−3)(3x+2)