Question

without actual division show that p(x) =4x³+3x²-4x-3 is exactly divisible by (x+1) and (4x+3)

Answer 1

$x + 1 = 0 \\ x = - 1 \\ putting \: the \: value \: of \: x \\ p( - 1) = 4 {( - 1)}^{3} + 3 {( - 1)}^{2} - \\ 4( - 1) - \ 3 \\ = - 4 + 3 + 4 - 3 \\ = 0 \\ by \: factor \: theorem \\ p( - 1) = 0 \\ hence \:p(x) \: is \: \\ divisible \: by \: (x + 1)$
$4x + 3 = 0 \\ x = \frac{ - 3}{4} \\ putting \: value \: of \: x \: \\ p( \frac{ - 3}{4} ) = 4 \times ( - \frac{27}{64} ) + \\ 3 \times \frac{9}{16} + 3 - 3 \\ = \frac{ - 27}{16 } + \frac{27}{16} + 3 - 3 \\ = 0 \\ p(x) = 0 \\ by \: factor \:t heorem \: p(x) \: i s \: \\ divisible \: by \: (4x - 3) \\ \\ hence \: p(x) \: is \: divisible \: by \: \\ (x + 1) \: and \: (4x - 3) \\ - - - - - - - - - - - \\ hope \: it \: helps \: u \: \\ mark \: \: as \: brainliest$

Answer 2

Answer:

\begin{gathered}x + 1 = 0 \\ x = - 1 \\ putting \: the \: value \: of \: x \\ p( - 1) = 4 {( - 1)}^{3} + 3 {( - 1)}^{2} - \\ 4( - 1) - \ 3 \\ = - 4 + 3 + 4 - 3 \\ = 0 \\ by \: factor \: theorem \\ p( - 1) = 0 \\ hence \:p(x) \: is \: \\ divisible \: by \: (x + 1) \\ \end{gathered}

x+1=0

x=−1

puttingthevalueofx

p(−1)=4(−1)

3

+3(−1)

2

−

4(−1)− 3

=−4+3+4−3

=0

byfactortheorem

p(−1)=0

hencep(x)is

divisibleby(x+1)

\begin{gathered}4x + 3 = 0 \\ x = \frac{ - 3}{4} \\ putting \: value \: of \: x \: \\ p( \frac{ - 3}{4} ) = 4 \times ( - \frac{27}{64} ) + \\ 3 \times \frac{9}{16} + 3 - 3 \\ = \frac{ - 27}{16 } + \frac{27}{16} + 3 - 3 \\ = 0 \\ p(x) = 0 \\ by \: factor \:t heorem \: p(x) \: i s \: \\ divisible \: by \: (4x - 3) \\ \\ hence \: p(x) \: is \: divisible \: by \: \\ (x + 1) \: and \: (4x - 3) \\ - - - - - - - - - - - \\ hope \: it \: helps \: u \: \\ mark \: \: as \: brainliest\end{gathered}

4x+3=0

x=

4

−3

puttingvalueofx

p(

4

−3

)=4×(−

64

27

)+

3×

16

9

+3−3

=

16

−27

+

16

27

+3−3

=0

p(x)=0

byfactortheoremp(x)is

divisibleby(4x−3)

hencep(x)isdivisibleby

(x+1)and(4x−3)

−−−−−−−−−−−

hopeithelpsu

markasbrainliest