for ax2+bx+c=10x2−7, find the values of a, b and c a=10 b=0 c=−7
for (2a−b)x2+(a+2b)x+8=4x2−3x+8, find values of a and b 2a−b=4 a+2b=−3 a=−3−2b −6−5b=4 b=−2 a=1
for (2a−3b)x2+(3a+b)x+c=7x2+5x+7, find the values of a, b and c 2a−3b=7 3a+b=5 c=7 b=5−3a 2a−15+9a=7
for 2x2+4x+5=a(x+b)2+c, find the values of a, b and c 2x2+4x+5=a(x2+2bx+b2)+c 2x2+4x+5=ax2+2abx+ab2+c a=2 2ab=4 ab2+c=5 b=1 c=3
express x2 in the form c(x+2)2+a(x+2)+d x2=c(x2+4x+4)+ax+2a+d x2=cx2+4cx+4c+ax+2a+d x2=cx2+(4c+a)x+4c+2a+d c=1 4c+a=0 4c+2a+d=0 a=−4 d=4
so the answer is 1(x+2)2+−4(x+2)+4
express x3 in the form (x+1)3+a(x+1)2+b(x+1)+c x3=x3+3x2+3x+1+a(x2+2x+1)+bx+b+c x3=x3+
find the values of a, b and c such that x2=a(x+1)2+bx+c =a(x2+2x+1)+bx+c =ax2+2ax+a+bx+c =ax2+(2a+b)x+a+c
a=1
2a+b=0
b=-2
a+c=0
c=-1
show that 3x3−9x2+8x+2 cannot be expressed in the form a(x+b)3+c a(x+b)3+c=a(x3+3x2b+3xb2+b3)+c =ax3+3ax2b+3axb2+ab3+c a=3 ab3+c=2 3ab2=8 3ab=−9 b=−1
See that 3ab2=8 is false. So cannot.
if 3x2−9x2+9x+2 can be expressed in the form a(x+b)3+c, then find the values of a, b and c a(x+b)3+c=a(x3+3x2b+3xb3+b3)+c =ax3+3ax2b+3axb2+ab3+c a=3 3ab=−9 3ab2=9 ab3+c=2 b=−1 −3+c=2 c=5
show that constants a, b, c, and d can be found such that n3=a(n+1)(n+2)(n+3)+b(n+1)(n+2)+c(n+1)+d n3=a(n+1)(n2+5n+6)+bn(n2+3n+2)+cn+c+d n3=a+an3+a5n2+a6n+an2+a5n+a6+bn3+b3n2+b2n+cn+c+d n3=an3+(6a+b)n2+(11a+3b+c)n+6a+2b+c+d a=1 6a+b=0 b=−6 11a+3b+c=0 c=7 6a+2b+c+d=0 d=−1
show that no constants a and b can be found such that n2=a(n+1)(n+2)+b(n+1)(n+2)+c(n+1)+d
express n2 in the form a(n+1)(n+2)+b(n+1)+c
express a(x+b)2+c in expanded form
express ax2+bx+c in completed square form
prove that, if ax3+bx2+cx+d=(x+1)2(px+q), then b=d−2a and c=a−2d
if 3x2+10x+3=c(x−a)(x−b) for all values of x, find the values of a, b and c
for any number n, show that n2 can be expressed as a(n−1)2+b(n−2)2+c(n−3)2, and find the values of a, b and c
if x3+2x2−9x+c can be expressed in the form (x−a)2(x−b), show that either c=5 or c=−27, and find a and b for each of these cases.
A polynomial P is said to be even if P(−x)=P(x) for all x. A polynomial P is said to be odd if P(−x)=−P(x) for all x.
show that, if P(x)=ax4+bx3+cx2+dx+e is even, then b=d=0
show that, if P(x)=ax5+bx4+cx3+dx2+ex+f is odd, then b=d=f=0