Worked example

Finding q so that (1+qx)(1+3x)⁶ has no term in x⁴

A-Level Maths · 97094 markseasy

Question

Find the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $(1 + 3 x)^{6}$ . It is given that, when $(1 + q x) (1 + 3 x)^{6}$ is expanded, there is no term in $x^{4}$ . Find the value of the constant $q$ .

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Full worked solution

(i) Coefficient of $x^{3}$ in $(1 + 3 x)^{6}$
We use the binomial theorem for $(a + b)^{n}$ , where the general term is $(r n) a^{n - r} b^{r}$ . For $(1 + 3 x)^{6}$ , $a = 1$ , $b = 3 x$ , $n = 6$ , so the term with $x^{3}$ occurs when $r = 3$ :
$\text{Term for } x^3 = \binom{6}{3} \cdot 1^{6-3} \cdot (3x)^3$
Substitute values: $(3 6) = 20$ , $(3 x)^{3} = 27 x^{3}$ , so the coefficient is $20 \cdot 27 = 540$ . Therefore the coefficient of $x^{3}$ is $540$ .
Note.
Binomial coefficient shortcut: $(r n) = \frac{n !}{r ! ( n - r )!}$ ; for $r = 3$ , $(3 6) = 20$ is a standard value you can recall.
(i) Coefficient of $x^{4}$ in $(1 + 3 x)^{6}$
Again use the binomial theorem, now with $r = 4$ to get the $x^{4}$ term:
$\text{Term for } x^4 = \binom{6}{4} \cdot 1^{6-4} \cdot (3x)^4$
Substitute values: $(4 6) = (2 6) = 15$ , $(3 x)^{4} = 81 x^{4}$ , so the coefficient is $15 \cdot 81 = 1215$ . Therefore the coefficient of $x^{4}$ is $1215$ .
Note.
$(r n) = (n - r n)$ , so $(4 6) = (2 6)$ simplifies calculation to avoid factorials for larger $r$ .
(ii) Find $q$ such that there is no $x^{4}$ term in $(1 + q x) (1 + 3 x)^{6}$
When expanding $(1 + q x) (1 + 3 x)^{6}$ , the $x^{4}$ term comes from two products:
1. $1$ multiplied by the $x^{4}$ term of $(1 + 3 x)^{6}$ , with coefficient $1215$
2. $q x$ multiplied by the $x^{3}$ term of $(1 + 3 x)^{6}$ , with coefficient $q \cdot 540$
For no $x^{4}$ term, the total coefficient must equal zero:
$1215 + 540q = 0$
Rearrange to solve for $q$ : $q = - \frac{1215}{540} = - \frac{9}{4} = - 2.25$ . Therefore $q = - \frac{9}{4}$ .
Note.
Only combine terms that produce the power of $x$ you care about; ignore all other terms to save time.

Answer

(i) Coefficient of $x^{3} = 540$ , coefficient of $x^{4} = 1215$ ; (ii) $q = - \frac{9}{4}$

What this tests

Binomial expansion of positive integer powers
Binomial coefficient properties
Polynomial multiplication and term matching
Eliminating a term by setting its total coefficient to zero

⚠️

⚠ A common mistake is to multiply $q x$ by the $x^{4}$ term instead of the $x^{3}$ term, leading to an incorrect $x^{5}$ contribution. Always match exponents: $x^{1} \cdot x^{k} = x^{k + 1}$ , so to get $x^{4}$ , $k = 3$ .

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Finding q so that (1+qx)(1+3x)⁶ has no term in x⁴

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