The polynomial is (mx^3+3)(2x²+5x+2)-(8x^5 +20x^4)
if it is reduced to 8x^3+6x²+15x+6, so we can find the value of m
(mx^3+3)(2x²+5x+2)-(8x^5+20x^4) = <span>8x^3+6x²+15x+6
</span>2mx^5+5mx^4+2mx^3+6x²+15x+6-8x^5-20x^4=<span>8x^3+6x²+15x+6
</span>2mx^5+5mx^4+2mx^3=8x^3+6x²+15x+6-6x²-15x-6+ <span>8x^5+20x^4
</span>= 8x^5+20x^4+<span>8x^3= 4(2x^5+5x^4+2x^3)
finally
</span>m(2x^5+5x^4+2x^3)=<span>4(2x^5+5x^4+2x^3), and after simplification
</span>
C: m=4
<span>4. When the expression is factored x²-3x-18 completely,
</span>
one of its factor is x-6
<span>x²-3x-18=0
</span>D= 9-4(-18)= 81, sqrtD=9 x=3-9/2= -6/2= -3, and x=3+9 / 2= 6
so <span>x²-3x-18= (x-6)(x+6)
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1) 1,2,3,4,5,6,7
2) -3,-2,-1,0,1,2,3,4,5,6,7
3)-3,-2,-1
4)1,2,3,4
(3*j)-8
I thinkkkk... I’m positive tho
Answer:
Since there are six possible outcomes, the probability is 1/6.
