QUESTION 11
Easy
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
1
Use the given conditions to find \(a, b, c,\) and \(d\).
(i) \(p(0) = 5\) ⇒ \(a(0) + b = 5\) ⇒ \(b = 0\)
(iii) \(p(x) + q(x) = 6x + 4\) ⇒ \( (a + c)x + (b + d) = 6x + 4\)
So \(a + c = 6\), and
\(b + d = 4 ⇒ 5 + d = 4 ⇒ d = −1\)
(ii) \(p(x) − q(x)\) cuts \(x-\)axis at (\(3,0\)): \(p(3) − q(3) = 0\)
\( (3a + 5) − (3c − 1) = 0\)
⇒ \(3a + 5 − 3c + 1 = 0\)
⇒ \(3a − 3c + 6 = 0\)
⇒ \(3(a − c) = −6\)
⇒ \(a − c = −2\)
Solving \(a + c = 6\) and \(a − c = −2\), we get
\(2a = 4\) ⇒ \(a = 2\). Then \(c = 4\).
(iii) \(p(x) + q(x) = 6x + 4\) ⇒ \( (a + c)x + (b + d) = 6x + 4\)
So \(a + c = 6\), and
\(b + d = 4 ⇒ 5 + d = 4 ⇒ d = −1\)
(ii) \(p(x) − q(x)\) cuts \(x-\)axis at (\(3,0\)): \(p(3) − q(3) = 0\)
\( (3a + 5) − (3c − 1) = 0\)
⇒ \(3a + 5 − 3c + 1 = 0\)
⇒ \(3a − 3c + 6 = 0\)
⇒ \(3(a − c) = −6\)
⇒ \(a − c = −2\)
Solving \(a + c = 6\) and \(a − c = −2\), we get
\(2a = 4\) ⇒ \(a = 2\). Then \(c = 4\).
2
Substitute the values to find \(p(x)\) and \(q(x)\)
\(p(x) = 2x + 5\) and \(q(x) = 4x − 1\).
🏆
Final Answer : \(p(x) = 2x + 5\), \(q(x) = 4x − 1\)
Concept Note
A linear polynomial is of the form \(p(x) = mx + c\), where \(m\) is the slope and \(c\) is the \(y-\)intercept.
The problem uses basic polynomial identity and the given conditions to find the polynomials.