Question

without actual division prove that 3x^4 - 2x^3 - 2x^2 - 2x - 5​

Answer 1

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Step by step solution:

$Given \ p(x)=3x^4 - 2x^3 - 2x^2 - 2x - 5\\\\and \ g(x)=3x^{2}-2x-5\\\\we \ have \ to \ prove \ p(x) \ is \ exactly \ divisible \ by \ g(x)\\\\First \ let \ us \ find \ zeroes \ of \ g(x)\\\\g(x)=3x^{2}-2x-5\\\\g(x)=3x^{2}-5x+3x-5\\\\g(x)=x(3x-5)+(3x-5)\\\\g(x)=(3x-5)(x+1)\\\\x=\frac{5}{3} \ or \ x=-1\\\\Now \ putting \ g(x) \ in \ p(x) \ to \ check$

$p(x)=3x^4 - 2x^3 - 2x^2 - 2x - 5\\\\p(-1)=3 \times (-1)^4-2\times (-1)^3-2 \times (-1)^2-2(-1)-5\\\\p(-1)=3+2-2+2+5\\\\p(-1)=0\\\\Now \ putting \ x=\frac{5}{3}\\\\p(\frac{5}{3})=3 \times(\frac{5}{3})^4-2\times(\frac{5}{3})^3-2\times(\frac{5}{3})^2-2\times(\frac{5}{3})-5$

$p(\frac{5}{3})=\frac{1875}{81}-\frac{250}{27}-\frac{50}{9}-\frac{10}{3}-5\\\\p(\frac{5}{3})=\frac{1875-750-450-270-405}{81}\\\\p(\frac{5}{3})=\frac{1875-1875}{81}\\\\p(\frac{5}{3})=0\\\\Both \ are \ factors \ of \ p(x)\\\\Hence \ p(x) \ is \ exactly \ divisible \ by \ g(x)$

Hope it is clear to you.