Question
Factorize:
$9(2a - b)^2 - 4(2a - b) - 13$
$9(2a - b)^2 - 4(2a - b) - 13$
✓
Answer
Let $2a - b = x = 9x^2 - 4x - 13$
Splitting the middle term,
$= 9x^2 - 13x + 9x - 13$
$= x(9x - 13) + 1(9x - 13)$
$= (9x - 13)(x + 1)$
Substituting$ x = 2a - b = [9(2a - b) - 13](2a - b + 1) $
$= (18a - 9b - 13)(2a - b + 1)$
$\therefore$ $9(2a - b)^2 - 4(2a - b) - 13$
$= (18a - 9b - 13)(2a - b + 1)$
Splitting the middle term,
$= 9x^2 - 13x + 9x - 13$
$= x(9x - 13) + 1(9x - 13)$
$= (9x - 13)(x + 1)$
Substituting$ x = 2a - b = [9(2a - b) - 13](2a - b + 1) $
$= (18a - 9b - 13)(2a - b + 1)$
$\therefore$ $9(2a - b)^2 - 4(2a - b) - 13$
$= (18a - 9b - 13)(2a - b + 1)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The radius and height of a right circular cone are in the ratio $5 : 12$ and its volume is $2512$ cubic cm. Find the slant height and radius of the cone. $($Use $\pi=3.14).$
→If ABCD is a cyclic quadrilateral in which AD || BC. Prove that $\angle\text{B}=\angle\text{C}.$
→Construct a histogram for the following data:
| Monthly School fee (in ₹): | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 | 210-240 |
| No of schools | 5 | 12 | 14 | 18 | 10 | 9 | 4 |
In the given figure, AB || CD transversal t cuts them at E and F respectively. If EG and FG are the bisectors of $\angle\text{BEF}$ and $\angle\text{EFD}$ respectively, prove that $\angle\text{EGF}=90^\circ.$
→Factorize:
$a^2 + 4b^2 - 4ab - 4c^2$
→$a^2 + 4b^2 - 4ab - 4c^2$
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by:$5 + 2x$
→Find the values of x in each of the following:$5^{\text{x}-2}\times3^{2\text{x}-3}=135$
→Find the value k if $x - 3$ is a factor of $k^2x^3$- $kx^2$ + $3kx$ -$k.$
→Write two solutions for the following equations:
x = 6y
x = 6y
Find three rational numbers lying between $\frac{3}{5}$ and $\frac{7}{8}.$ How many rational numbers can be determined between these two numbers?
→