Divide 23x^2‒2x^4‒4x^3+12+x^5‒31x by x^3‒7 x+5 then reminder
Answers
Answer:
The given polynomial is p(x)=x
4
−2x
3
+3x
2
−ax+3a−7
Given that, the polynomial p(x) when divided by (x+1) leaves remainder 19
Therefore, p(−1)=19 (By Remainder theorem)
=>(−1)
4
−2×(−1)
3
+3(−1)
2
−(−1)a+3a−7=19
=>1+2+3+a+3a−7=19
=>4a−1=19
=>4a=20
=>a=5
The value of a is 5
Now,
p(x)=x
4
−2x
3
+3x
2
−5x+3×5−7
=x
4
−2x
3
+3x
2
−5x+15−7
=x
4
−2x
3
+3x
2
−5x+8
Remainder when the polynomial is divided by (x+2)
=p(−2) (By Remainder Theorem)
=−2
4
−2(−2)
3
+3(−2)
2
−5(−2)+8
=16+16+12+10+8
=62
Thus, the remainder of the polynomial p(x) when divided by (x+2) is 62
Given that, the polynomial p(x) when divided by (x+1) leaves remainder 19
Therefore, p(−1)=19 (By Remainder theorem)
=>(−1)
4
−2×(−1)
3
+3(−1)
2
−(−1)a+3a−7=19
=>1+2+3+a+3a−7=19
=>4a−1=19
=>4a=20
=>a=5
The value of a is 5
Now,
p(x)=x
4
−2x
3
+3x
2
−5x+3×5−7
=x
4
−2x
3
+3x
2
−5x+15−7
=x
4
−2x
3
+3x
2
−5x+8
Remainder when the polynomial is divided by (x+2)
=p(−2) (By Remainder Theorem)
=−2
4
−2(−2)
3
+3(−2)
2
−5(−2)+8
=16+16+12+10+8
=62
Thus, the remainder of the polynomial p(x) when divided by (x+2) is 62