Question

21. (a) If f(x) = 24x+px? -5x +q has two factors 2x +1 and 3x -1 then find p and q. Also, factorize
f(x) completely.
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PLEASE SOLVE IT VER FAST

BECAUSE

IT'S SO URGENT.​

Answer 1

Given:-

• f(x) = 24x³ + px² - 5x + q

• g(x) = 2x + 1

• g(x) = 3x - 1

To Find :-

• The Value of P and Q

Now,

→ g(x) = 2x + 1 = 0

→ 2x = - 1

→ x = -1/2

Putting the value of "x" in f(x).

→ f (x) = 24x³ + px² - 5x + q

→ 24 ( -1/2)³ + p ( -1/2)² - 5 ( -1/2) + q = 0

→ $\sf{24 * \dfrac{ -1}{8} + p * \dfrac{1}{4} + 5 * \dfrac{1}{2} + q = 0}$

→ $\sf{ -3 + \dfrac{P}{4} + \dfrac{5}{2} + q = 0}$

→ $\sf{\dfrac{ -12 + P + 10 + 4q}{ 4 } = 0}$

→ $\sf{ -12 + P + 10 + 4q = 0}$

→ $\sf{ -2 + P + 4q = 0}$

→ $\sf{ P + 4q = 2 }$..........1

Again,

→ g ( x ) = 3x - 1 = 0

→ 3x = 1

→ x = 1/3

Putting the Value of "x" in f(x)

→ f (x) = 24x³ + px² - 5x + q

→ 24 ( 1/3)³ + p ( 1/3)² - 5 ( 1/3) + q

→ $\sf{24 * \dfrac{ 1}{27} + p * \dfrac{1}{9} + 5 * \dfrac{1}{3} + q }$

→ $\sf{ \dfrac{24}{27} + \dfrac{P}{9} + \dfrac{ 5}{3} + q = 0 }$

→ $\sf{ \dfrac{ 24 + 3P + 45 + 27q}{27}= 0}$

→ $\sf{ 3 ( 8 + P + 15 + 9q ) = 0/3}$

→ $\sf{ 8 + P + 15 + 9q = 0}$

→ $\sf { 23 + P + 9q = 0}$

→ $\sf { p + 9q = -23 }$.......2

Subtracting eq. 1 and eq. 2

→ P + 9q - ( P + 4q ) = -23 - 2

→ P + 9q - P - 4q = -25

→ 5q = -25

→ q = -25/5.

→ q = -5

Now, Putting the value of q in eq. 1

→ P + 4q = 2

→ P + 4 × -5 = 2

→ P- 20 = 2

→ P = 2 + 20

→ P = 22

Hence, The Value of P and Q is 22 and -5

Answer 2

$\huge \bf \star \{ \pink q \blue u \red e \purple s \green t \pink i \blue o \red n \}$

(a) If f(x) = 24x+px? -5x +q has two factors 2x +1 and 3x -1 then find p and q. Also, factorize

f(x) completely.

$\large \bf given$

$f(x) = 24x³ + px² - 5x + q \\ </p><p>g(x) = 2x + 1 \\ </p><p>g(x) = 3x$

$\large \bf \: to \: find$

$The \: Value \: of \: P \: and \: Q.$

$\large \ \bf \: solution$

$g(x) = 2x + 1 = 0 \\ </p><p></p><p>2x = - 1 \\ </p><p></p><p> x = -1/2 \\ </p><p>$

put the value of x in f(x)

$f (x) = 24x³ + px² - 5x + q \\ </p><p></p><p> 24 ( -1/2)³ + p ( -1/2)² - 5 ( -1/2) + q = 0</p><p>$

$→ \sf{24 \times \dfrac{ -1}{8} + p \times \dfrac{1}{4} + 5 \times \dfrac{1}{2} + q = 0}$

$\sf{ -3 + \dfrac{P}{4} + \dfrac{5}{2} + q = 0} \\ </p><p></p><p></p><p> \sf{\dfrac{ -12 + P + 10 + 4q}{ 4 } = 0} \\ </p><p></p><p></p><p> \sf{ -12 + P + 10 + 4q = 0} \\ </p><p></p><p> \sf{ -2 + P + 4q = 0} \\ </p><p></p><p> \sf{ P + 4q = 2 }..........(1)equation</p><p>$

similarly

$g ( x ) = 3x - 1 = 0 \\ </p><p></p><p>3x = 1 \\ </p><p></p><p> x = 1/3</p><p>$

$\sf \star \pink Putting \: the \: Value \: of \: "x" \: in \: f(x)$

$f (x) = 24x³ + px² - 5x + q</p><p>$

$24 ( 1/3)³ + p ( 1/3)² - 5 ( 1/3) + q$

$\sf{24 \times \dfrac{ 1}{27} + p \times \dfrac{1}{9} + 5 \times \dfrac{1}{3} + q } \\ </p><p></p><p>\sf{ \dfrac{24}{27} + \dfrac{P}{9} + \dfrac{ 5}{3} + q = 0 } </p><p></p><p>$

$→\sf{ \dfrac{ 24 + 3P + 45 + 27q}{27}= 0}$

$\sf{ 3 ( 8 + P + 15 + 9q ) = 0} \\ </p><p></p><p> \sf{ 8 + P + 15 + 9q = 0} \\ </p><p></p><p>\sf { 23 + P + 9q = 0} \\ </p><p></p><p> \sf { p + 9q = -23 } ......(.2)equation</p><p>$

$\sf \pink\bf \: Subtracting \: eq. 1 \: from \: eq. 2$

$P + 9q - ( P + 4q ) = -23 - 2 \\ </p><p></p><p> P + 9q - P - 4q = -25 \\ </p><p></p><p> 5q = -25 \\ </p><p></p><p>q = -25/5. \\ </p><p></p><p>q = -5</p><p>$

Now, Putting the value of q in eq. 1

$P + 4q = 2 \\ </p><p></p><p>P + 4 × -5 = 2 \\ </p><p></p><p> P- 20 = 2 \\ </p><p></p><p> P = 2 + 20 \\ </p><p></p><p>P = 22 \\ </p><p>$

$\bf \small \blue \star \: Hence, \: The \: Value \: of \: P \ \: Q \: is \: 22 \ and \: -5$