11. Let and be two linear polynomials such that: (i) . (ii) The polynomial cuts the -axis at . (iii) The sum is equal to for all real . Find the polynomials and .

Understand the Question We are given two linear polynomials, and , with four unknown constants: and . We use each given condition to set up algebraic equations: 1. directly gives the constant term . 2. allows us to compare coefficients for and the constant term to find relations between and determine . 3. cutting the -axis at means that at , the value of . Solving the resulting linear system gives the coefficients and , completely determining both polynomials. Step 1 · Find the constant terms and Given . Using condition (i): Now, write the sum of and : Given that for all real , compare corresponding coefficients: Substitute into : Step 2 · Use condition (ii) to form a second equation in and Find the expression for : Since cuts the -axis at , we have : Step 3 · Solve for and state the polynomials Add equations and : Substitute into equation : Substitute the values of into and : Answer > and Common Mistakes -intercept Interpretation: Forgetting that cutting the -axis at means the polynomial evaluates to at , i.e., . Sign Errors in Subtraction: Making a sign error when subtracting negative terms: , not .

Introduction to Linear Polynomials | EOT

Question 11

*11. Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that:

(i) $p(0) = 5$ .

(ii) The polynomial $p(x) - q(x)$ cuts the $x$ -axis at $(3, 0)$ .

(iii) The sum $p(x) + q(x)$ is equal to $6x + 4$ for all real $x$ .

Find the polynomials $p(x)$ and $q(x)$ .

Solution

Understand the Question

We are given two linear polynomials, $p(x) = ax + b$ and $q(x) = cx + d$ , with four unknown constants: $a, b, c,$ and $d$ .
We use each given condition to set up algebraic equations:
1. $p(0) = 5$ directly gives the constant term $b$ .
2. $p(x) + q(x) = 6x + 4$ allows us to compare coefficients for $x$ and the constant term to find relations between $a, c$ and determine $d$ .
3. $p(x) - q(x)$ cutting the $x$ -axis at $(3, 0)$ means that at $x = 3$ , the value of $p(3) - q(3) = 0$ .
Solving the resulting linear system gives the coefficients $a, b, c,$ and $d$ , completely determining both polynomials.

Step 1 · Find the constant terms $b$ and $d$

Given $p(x) = ax + b$ .

Using condition (i):

p(0) = a(0) + b = 5 \implies b = 5

Now, write the sum of $p(x)$ and $q(x)$ :

\begin{aligned} p(x) + q(x) &= (ax + b) + (cx + d) \\ &= (a + c)x + (b + d) \end{aligned}

Given that $p(x) + q(x) = 6x + 4$ for all real $x$ , compare corresponding coefficients:

a + c = 6 \quad \dots (1)

b + d = 4

Substitute $b = 5$ into $b + d = 4$ :

5 + d = 4 \implies d = -1

Step 2 · Use condition (ii) to form a second equation in $a$ and $c$

Find the expression for $p(x) - q(x)$ :

\begin{aligned} p(x) - q(x) &= (ax + 5) - (cx - 1) \\ &= (a - c)x + 5 - (-1) \\ &= (a - c)x + 6 \end{aligned}

Since $p(x) - q(x)$ cuts the $x$ -axis at $(3, 0)$ , we have $(p - q)(3) = 0$ :

\begin{aligned} (a - c)(3) + 6 &= 0 \\ 3(a - c) &= -6 \\ a - c &= -2 \quad \dots (2) \end{aligned}

Step 3 · Solve for $a, c$ and state the polynomials

Add equations $(1)$ and $(2)$ :

\begin{aligned} (a + c) + (a - c) &= 6 + (-2) \\ 2a &= 4 \\ a &= 2 \end{aligned}

Substitute $a = 2$ into equation $(1)$ :

2 + c = 6 \implies c = 4

Substitute the values of $a, b, c, d$ into $p(x)$ and $q(x)$ :

p(x) = 2x + 5

q(x) = 4x - 1

Answer

$p(x) = 2x + 5$ and $q(x) = 4x - 1$

Common Mistakes

$x$ -intercept Interpretation: Forgetting that cutting the $x$ -axis at $(3, 0)$ means the polynomial evaluates to $0$ at $x = 3$ , i.e., $p(3) - q(3) = 0$ .
Sign Errors in Subtraction: Making a sign error when subtracting negative terms: $(ax + 5) - (cx - 1) = (a - c)x + 6$ , not $(a - c)x + 4$ .

Question 11

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