问题1：求多项式(x)的值= 5x − 4x 2 + 3
(i)x = 0
(ii)x = –1
(iii)x = 2
解决方案：
Given equation: 5x − 4x2 + 3
Therefore, let f(x) = 5x – 4x2 + 3
(i) When x = 0
f(0) = 5(0)-4(0)2+3
= 3
(ii) When x = -1
f(x) = 5x−4x2+3
f(−1) = 5(−1)−4(−1)2+3
= −5–4+3
= −6
(iii) When x = 2
f(x) = 5x−4x2+3
f(2) = 5(2)−4(2)2+3
= 10–16+3
= −3
问题2：为以下每个多项式找到p(0)，p(1)和p(2)：
(i)p(y)= y 2 −y + 1
(ii)p(t)= 2 + t + 2t 2 -t 3
(iii)p(x)= x 3
(iv)P(x)=(x-1)(x + 1)
解决方案：
(i) p(y) = y2 – y + 1
Given equation: p(y) = y2–y+1
Therefore, p(0) = (0)2−(0)+1 = 1
p(1) = (1)2–(1)+1 = 1
p(2) = (2)2–(2)+1 = 3
Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y2–y+1
(ii) p(t) = 2 + t + 2t2 − t3
Given equation: p(t) = 2+t+2t2−t3
Therefore, p(0) = 2+0+2(0)2–(0)3 = 2
p(1) = 2+1+2(1)2–(1)3 = 2+1+2–1 = 4
p(2) = 2+2+2(2)2–(2)3 = 2+2+8–8 = 4
Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t2−t3
(iii) p(x) = x3
Given equation: p(x) = x3
Therefore, p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x3
(iv) p(x) = (x−1)(x+1)
Given equation: p(x) = (x–1)(x+1)
Therefore, p(0) = (0–1)(0+1) = (−1)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3
Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)
问题3：验证以下各项是否为多项式的零(相对于它们的指示)。
(i)p(x)= 3x + 1，x = −1 / 3
(ii)p(x)= 5x–π，x = 4/5
(iii)p(x)= x 2 -1，x = 1，-1
(iv)p(x)=(x + 1)(x–2)，x = −1，2
(v)p(x)= x 2 ，x = 0
(vi)p(x)= lx + m，x = −m / l
(vii)p(x)= 3x 2 -1，x = -1 /√3，2 /√3
(viii)p(x)= 2x + 1，x = 1/2
解决方案：
(i) p(x)=3x+1, x=−1/3
Given: p(x)=3x+1 and x=−1/3
Therefore, substituting the value of x in equation p(x), we get.
For, x = -1/3
p(−1/3) = 3(-1/3)+1
= −1+1
= 0
Hence, p(x) of -1/3 = 0
(ii) p(x)=5x–π, x = 4/5
Given: p(x)=5x–π and x = 4/5
Therefore, substituting the value of x in equation p(x), we get.
For, x = 4/5
p(4/5) = 5(4/5)–π
= 4–π
Hence, p(x) of 4/5 ≠ 0
(iii) p(x)=x2−1, x=1, −1
Given: p(x)=x2−1 and x=1, −1
Therefore, substituting the value of x in equation p(x), we get.
For x = 1
p(1) = 12−1
=1−1
= 0
For, x = -1
p(−1) = (-1)2−1
= 1−1
= 0
Hence, p(x) of 1 and -1 = 0
(iv) p(x) = (x+1)(x–2), x =−1, 2
Given: p(x) = (x+1)(x–2) and x =−1, 2
Therefore, substituting the value of x in equation p(x), we get.
For, x = −1
p(−1) = (−1+1)(−1–2)
= (0)(−3)
= 0
For, x = 2
p(2) = (2+1)(2–2)
= (3)(0)
= 0
Hence, p(x) of −1, 2 = 0
(v) p(x) = x2, x = 0
Given: p(x) = x2 and x = 0
Therefore, substituting the value of x in equation p(x), we get.
For, x = 0
p(0) = 02 = 0
Hence, p(x) of 0 = 0
(vi) p(x) = lx+m, x = −m/l
Given: p(x) = lx+m and x = −m/l
Therefore, substituting the value of x in equation p(x), we get.
For, x = −m/l
p(-m/l)= l(-m/l)+m
= −m+m
= 0
Hence, p(x) of -m/l = 0
(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3
Given: p(x) = 3x2−1 and x = -1/√3 , 2/√3
Therefore, substituting the value of x in equation p(x), we get.
For, x = -1/√3
p(-1/√3) = 3(-1/√3)2 -1
= 3(1/3)-1
= 1-1
= 0
For, x = 2/√3
p(2/√3) = 3(2/√3)2 -1
= 3(4/3)-1
= 4−1
=3 ≠ 0
Hence, p(x) of -1/√3 = 0
but, p(x) of 2/√3 ≠ 0
(viii) p(x) =2x+1, x = 1/2
Given: p(x) =2x+1 and x = 1/2
Therefore, substituting the value of x in equation p(x), we get.
For, x = 1/2
p(1/2) = 2(1/2)+1
= 1+1
= 2≠0
Hence, p(x) of 1/2 ≠ 0
问题4：在以下每种情况下，找到多项式的零：
(i)p(x)= x + 5
(ii)p(x)= x-5
(iii)p(x)= 2x + 5
(iv)p(x)= 3x–2
(vii)p(x)= cx + d，c≠0，c，d为实数。
解决方案：
(i) p(x) = x+5
Given: p(x) = x+5
To find the zero, let p(x) = 0
p(x) = x+5
0 = x+5
x = −5
Therefore, the zero of the polynomial p(x) = x+5 is when x = -5
(ii) p(x) = x–5
Given: p(x) = x–5
p(x) = x−5
x−5 = 0
x = 5
Therefore, the zero of the polynomial p(x) = x–5 is when x = 5
(iii) p(x) = 2x+5
Given: p(x) = 2x+5
p(x) = 2x+5
2x+5 = 0
2x = −5
x = -5/2
Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2
(iv) p(x) = 3x–2
Given: p(x) = 3x–2
p(x) = 3x–2
3x−2 = 0
3x = 2
x = 2/3
Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3
(v) p(x) = 3x
Given: p(x) = 3x
p(x) = 3x
3x = 0
x = 0
Therefore, the zero of the polynomial p(x) = 3x is when x = 0
(vi) p(x) = ax, a0
Given: p(x) = ax, a≠ 0
p(x) = ax
ax = 0
x = 0
Therefore, the zero of the polynomial p(x) = ax is when x = 0
(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.
Given: p(x) = cx+d
p(x) = cx + d
cx+d =0
x = -d/c
Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c