Question

question 4, 10

please solve it fast ​

Answer 1

Correct Question 4 :

Find the value of m , so that 2x - 1 be a factor of 8x³ + 4x³ - 16x² + 10x + m

Answer :

Let f( x ) = 8x^4 + 4x³ - 16x² + 10x + m

Given :

( 2x - 1 ) ⇒ ( x - 1/2 ) is a factor of f( x )

We know that :

By factor theorem :

If ( x - a ) is a factor of a polynomial f( x ) then f( a ) = 0

Hence, f( 1/2 ) = 0

⇒ f( 1/2 ) = 0

⇒ 8( 1/2 )^4 + 4( 1/2 )³ - 16( 1/2 )² + 10( 1/2 ) + m = 0

⇒ 8( 1/16 ) + 4( 1/8 ) - 16( 1/4 ) + 5 + m = 0

⇒ 1/2 + 1/2 - 4 + 5 + m = 0

⇒ 1 + 1 + m = 0

⇒ 2 + m = 0

⇒ m = - 2

Therefore the value of m is - 2.

Question 10 :

Factorize

(i) 1 + 64x³

It can be written as

= 1³ + ( 4x )³

Since a³ + b³ = ( a + b )( a² + b² - ab )

= ( 1 + 4x )[ 1² + ( 4x )² - 1( 4x ) ]

= ( 1 + 4x )( 1 + 16x² - 4x )

(ii) a³ - 2√2b³

It can be written as

= a³ - ( √2 b)³

Since x³ - y³ = ( x + y )( x² + y² + xy )

= ( a - √2b )[ a² + ( √2b )² + a( √2b ) ]

= ( a - √2b)( a² + 2b² + √2 ab)

Hence factorised.

Answer 2

★ Question (4)

find the value of m so that ( 2x-1 ) be a factor of 8x^4+4x^3-16x^2+10x+m

★ Solution

Putting (2x-1)=0

→ 2x=1

→ x=1/2

By remainder theorem

$\tt\rightarrow 8x^4+4x^3-16x^2+10x+m\\\tt\rightarrow 8(\frac{1}{2})^4+4(\frac{1}{2})^3-16(\frac{1}{2}^2+10(\frac{1}{2}+m\\\tt\rightarrow \frac{8}{16}+\frac{4}{8}-\frac{16}{4}+\frac{10}{2}+m=0\\\tt\rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0\\\tt\rightarrow \frac{1+1}{2}-4+5+m=0\\\tt\rightarrow 1-4+5+m=0\\\tt m= 2$

★ Question (10)

1st part

→ factorize $1+64x^3$

★identity used

$\tt{\red{\underline{\fbox{ (x^3+y^3)=(x+y)(x^2-ab+y^2)}}}}$

★ Solution

$\leadsto\tt 1+64^3\\\tt\leadsto 1^3+(4x)^3\\\tt\leadsto (1+4x)[1^2+(4x)^2-1(4x)]\\\tt\leadsto (1+4x)(1+16x^2-4x)$

2nd part

★identity used

$\tt{\pink{\underline{\fbox{ (x^3-y^3)=(x+y)(x^2+ab+y^2)}}}}$

$\tt\:factroize\:\:→ a^3-2\sqrt b^3 \\\tt\leadsto a^3(\sqrt2 b^3)\\\tt\leadsto (a-2\sqrt b)[(a^2+a(\sqrt2b)+(2\sqrt b^2)]\\\tt\leadsto (a-2\sqrt b)(a^2+\sqrt2 ab + 2b^2)$