- #1
Monoxdifly
MHB
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- 0
The remainder of \(\displaystyle p(x)=x^3+ax^2+4bx-1\) divided by \(\displaystyle x^2+1\) is –5a + 4b. If the remainder of p(x) divided by x + 1 is –a – 2, the value of 8ab is ...
A. \(\displaystyle -\frac34\)
B. \(\displaystyle -\frac12\)
C. 0
D. 1
E. 3
Dividing p(x) by \(\displaystyle x^2+1\) by \(\displaystyle x^2+1\) with –5a + 4b as the remainder using long division, I got (4bx – 1) – ((a – 1)x + a – 1) = –5a + 4b, thus (4b – a + 1)x + a = –5a + 4b. Does this mean that 4b – a + 1 = 0 since the right hand doesn't have an x term? Or do I need to look for the value of x first? I'm at a loss here.
A. \(\displaystyle -\frac34\)
B. \(\displaystyle -\frac12\)
C. 0
D. 1
E. 3
Dividing p(x) by \(\displaystyle x^2+1\) by \(\displaystyle x^2+1\) with –5a + 4b as the remainder using long division, I got (4bx – 1) – ((a – 1)x + a – 1) = –5a + 4b, thus (4b – a + 1)x + a = –5a + 4b. Does this mean that 4b – a + 1 = 0 since the right hand doesn't have an x term? Or do I need to look for the value of x first? I'm at a loss here.