Fill in the blanks to complete the following identities: Class 9

Fill in the blanks to complete the following identities: Class 9

Question 1.

Fill in the blanks to complete the following identities: Class 9

i. s² - 11s + 24 = (_______)(______)

ii. (______) (x + 1) = (3x² - 4x - 7)

iii. 10x² - 11x - 6 = (2x - ______)(______ + 2)

iv. 6x² + 7x + 2 = (_____)(_____)

Solution:

i. Comparing

s² - 11s + 24 with s² + (a + b)s + ab, we get

a + b = -11 and ab = 24 .

Since (- 8) + (- 3) = -11 and (- 8) × (- 3) = 24,

a = -8 and b = -3 or vice-versa

∴ s² - 11s + 24 = (s - 8)(s - 3)

ii. 3x² - 4x - 7 = 3(x² - $\frac{4}{3}-\frac{7}{3}$)

Comparing

x² - $\frac{4}{3}-\frac{7}{3}$ with x² + (a + b) x + ab, we get

a + b = $-\frac{4}{3}$ and ab = $-\frac{7}{3}$

Since ($-\frac{7}{3}$) + 1 = $-\frac{4}{3}$ and

$\left(-\frac{7}{3}\right) \times(1)=\left(-\frac{7}{3}\right)$

a = $-\frac{7}{3}$ and b = 1 or vice-versa

∴ 3x² - 4x - 7 = 3 (x - $\frac{7}{3}$) (x + 1)

= (3x - 7)(x + 1)

∴ (3x - 7)(x + 1) = 3x² - 4x - 7

iii. 10x² - 11x - 6 = 10 (x² - $\frac{11}{10} x-\frac{6}{10}$)

= 10 (x² - $\frac{11}{10} x-\frac{3}{5}$)

Comparing

x² - $\frac{11}{10} x-\frac{3}{5}$ with x² + (a + b)x + ab, we get

a + b = $-\frac{11}{10}$ and ab = $-\frac{3}{5}$

Since $\left(-\frac{3}{2}\right)+\frac{2}{5}=-\frac{11}{10}$ and $\left(-\frac{3}{2}\right) \times\left(\frac{2}{5}\right)=-\frac{3}{5}$,

$\begin{array}{ll} & a=-\frac{3}{2} \text { and } b=\frac{2}{5} \text { or vice-versa } \\ \therefore & 10 x^2-11 x-6=10\left(x-\frac{3}{2}\right)\left(x+\frac{2}{5}\right)\end{array}$

= (2x - 3)(5x + 2)

iv. 6x² + 7x + 2 = 6 (x² + $\frac{7}{6} x+\frac{2}{6}$)

= 6(x² + $\frac{7}{6} x+\frac{1}{3}$)

Comparing

x² + $\frac{7}{6} x+\frac{1}{3}$ with x² + (a + b)x + ab, we get

a + b = $\frac{7}{6}$ and ab = $\frac{1}{3}$

$\begin{array}{ll} & \text { Since } \frac{1}{2}+\frac{2}{3}=\frac{7}{6} \text { and }\left(\frac{1}{2}\right) \times\left(\frac{2}{3}\right)=\frac{1}{3}, \\ & a=\frac{1}{2} \text { and } b=\frac{2}{3} \text { or vice-versa } \\ \therefore \quad & 6 x^2+7 x+2=6\left(x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\end{array}$

= (2x + 1)(3x + 2)