factorise 4x^4 -24x^3 +31x^2+6x-8=0
Answers
Answer:
f(x) = 4x⁴-24x³+31x²+6x-8 = 0
*Remember take factors of constant term.
factors of 8 are ± 1,± 2,±4,±8
putting,
x= 1
f(1) = 4(1)⁴-24(1)³+31(1)²+6(1)-8
= 4-24+31+6-8
= 4+6+31-24-8
= 41-32
= 9
therefore x-1 is not a factor.
x= -1
f(-1) = 4(-1)⁴-24(-1)³+31(-1)²+6(-1)-8
= 4+24+31-6-8
= 59-14
= 45
therefore x+1 is not a factor.
x= 2
f(2) = 4(2)⁴-24(2)³+31(2)²+6(2)-8
= 64 -192 +124 +12-8
= 200-200
= 0
★therefore x-2 is a factor of the polynomial.
(by factor theorem)
4x⁴-24x³+31x² +6x-8/ x-2
q (x)= 4x³-16x²-x+4
*For division refer the attachment.
factors of 4 are ± 1,±2,±4
putting
x = 1
q (1) = 4(1)³ - 16(1)² -(1) +4
= 4-16-1+4
= 8-17
= -9
x = -1
q(-1) = 4(-1)³ -16(-1)² -(-1) +4
= -4-16+1+4
= -20+5
= -15
x = 2
q(2) = 4(2)³ -16(2)²-(2)+4
= 32 -64 -2+4
= 36-66
= -30
x = -2
q(-2) = 4(-2)³-16(-2)² -(-2)+4
= -32 -64 +2+4
= -96+6
= -90
x = 4
q(4) = 4(4)³-16(4)²-(4)+4
= 256-256-4+4
= 0
★ therefore (x-4) is a factor of q(x).
q(x)/(x-4) = 4x²-1
* for divison refer the attachment
identity -
a²-b² = (a-b)(a+b)
4x²-1
4x² = (2x)²
1 = 1²
★(2x)² -(1)² = (2x-1)(2x+1)
★ therefore
4x⁴ -24x³+31 x² +6x-8
= (x-2)(x-4)(2x-1)(2x+1)
hence , factorized.
