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P2 2023 Oct Q4

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 4

题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

f(x)=4x3+ax2−29x+bf(x) = 4x^3 + ax^2 - 29x + b

where aa and bb are constants.

Given that (2x+1)(2x + 1) is a factor of f(x)f(x),

(a) show that a+4b=−56a + 4b = -56.

(2)

Given also that when f(x)f(x) is divided by (x−2)(x - 2) the remainder is −25-25,

(b) find a second simplified equation linking aa and bb.

(2)

(c) Hence, using algebra and showing your working,

(i) find the value of aa and the value of bb,

(ii) fully factorise f(x)f(x).

(5)
题目中文翻译

本题必须展示所有解题步骤,不能依赖计算器技术。

f(x)=4x3+ax2−29x+bf(x) = 4x^3 + ax^2 - 29x + b,其中 aa、bb 为常数。

已知 (2x+1)(2x + 1) 是 f(x)f(x) 的因式,

(a) 证明 a+4b=−56a + 4b = -56。

(2分)

又已知 f(x)f(x) 除以 (x−2)(x - 2) 的余数为 −25-25,

(b) 求另一个关于 aa 和 bb 的简化方程。

(2分)

(c) 由此,用代数方法并展示解题过程,

(i) 求 aa 和 bb 的值,

(ii) 完全因式分解 f(x)f(x)。

(5分)

解答

(a)

解法一

思路

展开

因式 (2x+1)(2x+1) 对应根 −12-\dfrac12。代入因式定理,再整理为题目要求的关系式。

答题过程

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Since (2x+1)(2x+1) is a factor, f(−1/2)=0f(-1/2)=0.

0= 4(−12)3+a(−12)2 −29(−12)+b= −12+a4+292+b= a4+14+b.\begin{align*} 0 =&\,4\bigg(-\frac12\bigg)^3 +a\bigg(-\frac12\bigg)^2\\ &\,\hspace{2pt}-29\bigg(-\frac12\bigg)+b\\ =&\,-\frac12+\frac a4+\frac{29}{2}+b\\ =&\,\frac a4+14+b. \end{align*}

Multiplying by 4:

a+4b=−56.(1)\boxed{a+4b=-56}.\qquad(1)

解法二

思路

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把已知因式与二次商相乘,比较各次项系数。先由首项、常数项、一次项确定商,再用二次项得到关系式。

答题过程

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Write

f(x)=(2x+1)(Ax2+Bx+C).f(x)=(2x+1)(Ax^2+Bx+C).

Comparing the coefficients of x3x^3, the constant term and xx gives

2A=4,C=b,2C+B=−29.2A=4,\quad C=b,\quad 2C+B=-29.

Thus A=2A=2 and B=−29−2bB=-29-2b. Comparing the coefficient of x2x^2:

a= A+2B= 2+2(−29−2b)= −56−4b.\begin{align*} a =&\,A+2B\\ =&\,2+2(-29-2b)\\ =&\,-56-4b. \end{align*}

Hence

a+4b=−56.\boxed{a+4b=-56}.

(b)

解法一

思路

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由余式定理,f(2)f(2) 等于题给余数 −25。

答题过程

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By the remainder theorem, f(2)=−25f(2)=-25:

4(2)3+a(2)2−29(2)+b= −2532+4a−58+b= −254a+b=1.(2)\begin{align*} 4(2)^3+a(2)^2-29(2)+b =&\,-25\\ 32+4a-58+b =&\,-25\\ \boxed{4a+b=1}.&\qquad(2) \end{align*}

解法二

思路

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直接做多项式长除法。依次减去三个商项对应的乘积,留下余数,再令其等于 −25。

答题过程

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Divide by x−2x-2. First subtract 4x2(x−2)4x^2(x-2):

f(x)−4x2(x−2)= (a+8)x2−29x+b.\begin{align*} f(x)-4x^2(x-2) =&\,(a+8)x^2-29x+b. \end{align*}

Next subtract (a+8)x(x−2)(a+8)x(x-2):

 (a+8)x2−29x+b −(a+8)x(x−2)= (2a−13)x+b.\begin{align*} &\,(a+8)x^2-29x+b\\ &\,\hspace{2pt}-(a+8)x(x-2)\\ =&\,(2a-13)x+b. \end{align*}

Finally subtract (2a−13)(x−2)(2a-13)(x-2):

 (2a−13)x+b−(2a−13)(x−2)= b+4a−26.\begin{align*} &\,(2a-13)x+b-(2a-13)(x-2)\\ =&\,b+4a-26. \end{align*}

This is the remainder, so

b+4a−26=−25⟹4a+b=1.b+4a-26=-25 \quad\Longrightarrow\quad \boxed{4a+b=1}.

(c)(i)

解法一

思路

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利用 (a)(b) 的两个关系式,消去 bb 后求 aa,再代回。

答题过程

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From (b), b=1−4ab=1-4a. Substitute into (a):

a+4(1−4a)= −56−15a+4= −56−15a= −60a= 4.\begin{align*} a+4(1-4a) =&\,-56\\ -15a+4 =&\,-56\\ -15a =&\,-60\\ a =&\,4. \end{align*}

Then b=1−4(4)=−15b=1-4(4)=-15, giving

a=4,b=−15.\boxed{a=4,\qquad b=-15}.

(c)(ii)

解法一

思路

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由此代入系数,并除以题给因式 2x+12x+1。补齐二次商的三步计算,再分解二次式。

答题过程

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Hence f(x)=4x3+4x2−29x−15f(x)=4x^3+4x^2-29x-15. Divide by the known factor 2x+12x+1.

Subtract the first quotient term, 2x22x^2:

f(x)−2x2(2x+1)=2x2−29x−15.f(x)-2x^2(2x+1)=2x^2-29x-15.

Subtract the next quotient term, xx:

 2x2−29x−15−x(2x+1)= −30x−15.\begin{align*} &\,2x^2-29x-15-x(2x+1)\\ =&\,-30x-15. \end{align*}

The final quotient term is −15-15, since −15(2x+1)=−30x−15-15(2x+1)=-30x-15. Thus the quotient is 2x2+x−152x^2+x-15.

Now factorise it:

2x2+x−15= 2x2+6x−5x−15= 2x(x+3)−5(x+3)= (2x−5)(x+3).\begin{align*} 2x^2+x-15 =&\,2x^2+6x-5x-15\\ =&\,2x(x+3)-5(x+3)\\ =&\,(2x-5)(x+3). \end{align*}

Therefore

f(x)=(2x+1)(2x−5)(x+3).\boxed{f(x)=(2x+1)(2x-5)(x+3)}.