Question

If the polynomial 2 x cube minus x square + bx + 4 x + 1 is a factor and leave reminder for when divided by 2 x + 1 find the value of a and b

Answer 1

Step-by-step explanation:

f(x)=2x

f(x)=2x 3

f(x)=2x 3 +ax

f(x)=2x 3 +ax 2

f(x)=2x 3 +ax 2 +bx−6

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 2

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−a=−8

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−a=−8a+b=4

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−a=−8a+b=4−8+b=4

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−a=−8a+b=4−8+b=4b=4+8=12

f(x)=2x 3 +ax 2 +bx−6(x−1) is a factor∴f(1)=0⇒2+a+b=6⇒a+b=4→1and f(x) leaves remainder 2 on being divided by (x−2)∴f(2)=2⇒2×2 3 +a×2 2 +b×2−6=2⇒4a+2b+16−6=2⇒4a+2b=2−10=−8⇒2a+b=−4Solving 1 and 22a+b=−4a+b=4−−−a=−8a+b=4−8+b=4b=4+8=12∴a=−8;b=12