check x^2+3x+7 is factor of 3x^4+5x^3+7x^2+2x+2
Answers
Answer:
Answer:
x^2+3x+1x
2
+3x+1 is a factor of 3x^4+5x^3-7x^2+2x+23x
4
+5x
3
−7x
2
+2x+2
Step-by-step explanation:
Divisor = x^2+3x+1x
2
+3x+1
Dividend = 3x^4+5x^3-7x^2+2x+23x
4
+5x
3
−7x
2
+2x+2
Now , Dividend = (Divisor \times Quotient)+RemainderDividend=(Divisor×Quotient)+Remainder
3x^4+5x^3-7x^2+2x+2= (x^2+3x+1 \times 3x^2)+(-4x^3-10x^2+2x+2)3x
4
+5x
3
−7x
2
+2x+2=(x
2
+3x+1×3x
2
)+(−4x
3
−10x
2
+2x+2)
3x^4+5x^3-7x^2+2x+2= (x^2+3x+1 \times 3x^2-4x)+(2x^2+6x+2)3x
4
+5x
3
−7x
2
+2x+2=(x
2
+3x+1×3x
2
−4x)+(2x
2
+6x+2)
3x^4+5x^3-7x^2+2x+2= (x^2+3x+1 \times 3x^2-4x+2)+(0)3x
4
+5x
3
−7x
2
+2x+2=(x
2
+3x+1×3x
2
−4x+2)+(0)
Thus remainder is 0
So, x^2+3x+1x
2
+3x+1 is a factor of 3x^4+5x^3-7x^2+2x+23x
4
+5x
3
−7x
2
+2x+2
Hence x^2+3x+1x
2
+3x+1 is a factor of 3x^4+5x^3-7x^2+2x+23x
4
+5x
3
−7x
2
+2x+2
Explanation:
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