Step-by-step explanation: {5x}^{2} – 29x + 205x 2 −29x+20 Factorise The Term By Middle Term Splitting. In Middle Term Splitting Divide Middle Term Such that On Multiplying Both It should be Product of Coffecient x² and constant Term \begin{gathered} {5x}^{2} – 29x + 20 = 0 \\ \\ {5x}^{2} – 25x – 4x + 20 = 0 \\ \\ 5x(x – 5) – 4(x – 5) = 0 \\ \\ (5x – 4)(x – 5) = 0\end{gathered} 5x 2 −29x+20=0 5x 2 −25x−4x+20=0 5x(x−5)−4(x−5)=0 (5x−4)(x−5)=0 Here One Zero of Polynomial \begin{gathered}5x – 4 = 0 \\ \\ 5x = 4 \\ \\ x = \frac{4}{5} \end{gathered} 5x−4=0 5x=4 x= 5 4 Other Zero of Polynomial \begin{gathered}x – 5 = 0 \\ \\ x = 5\end{gathered} x−5=0 x=5 We got Zeroes \begin{gathered} \alpha = \frac{4}{5} \\ \\ \beta = 5\end{gathered} α= 5 4 β=5 Polynomial ax²+bx+c {5x}^{2} – 29x + 205x 2 −29x+20 Coffecient of x²=5 Coffecient of x¹=-29 Coffecient of x^0=20 Let a in ax²+bx+c=5 Let B in ax²+bx+c=-29 Let c in ax²+bx+c=20 Sum Of Zeroes \begin{gathered} \alpha + \beta = – \frac{coffecient \: of \: x}{coffecient \: of \: {x}^{2} } \\ \\ \alpha + \beta = – \frac{b}{a} \\ \\ \frac{4}{5} + 5 = – \frac{-29}{5} \\ \\ \frac{29}{4} = + \frac{29}{4} < /p > < p > \end{gathered} α+β=− coffecientofx 2 coffecientofx α+β=− a b 5 4 +5=− 5 −29 4 29 =+ 4 29 </p><p> lhs = rhslhs=rhs Product of Zeroes \begin{gathered} \alpha \times \beta = \frac{constant \: term}{cofficient \: of \: {x}^{2} } \\ \\ \frac{4}{5} \times 5 = \frac{20}{5} \\ \\ \frac{20}{5} = \frac{20}{5} \\ \\ lhs = rhs\end{gathered} α×β= cofficientofx 2 constantterm 5 4 ×5= 5 20 5 20 = 5 20 lhs=rhs \boxed{\mathbf{\huge{LHS=RHS}}} LHS=RHS \underline{\mathbf{\huge{Verified✓}}} Verified✓ Reply
Answer: 5 and 4/5 Step-by-step explanation: 5x² – 29x + 20 =》 5x² -25x -4x + 20 =》 5x(x – 5) -4(x – 5) =》 (5x – 4) (x – 5) =》 x=4/5 and x = 5 Reply
Step-by-step explanation:
{5x}^{2} – 29x + 205x
2
−29x+20
Factorise The Term By Middle Term Splitting.
In Middle Term Splitting Divide Middle Term Such that On Multiplying Both It should be Product of Coffecient x² and constant Term
\begin{gathered} {5x}^{2} – 29x + 20 = 0 \\ \\ {5x}^{2} – 25x – 4x + 20 = 0 \\ \\ 5x(x – 5) – 4(x – 5) = 0 \\ \\ (5x – 4)(x – 5) = 0\end{gathered}
5x
2
−29x+20=0
5x
2
−25x−4x+20=0
5x(x−5)−4(x−5)=0
(5x−4)(x−5)=0
Here One Zero of Polynomial
\begin{gathered}5x – 4 = 0 \\ \\ 5x = 4 \\ \\ x = \frac{4}{5} \end{gathered}
5x−4=0
5x=4
x=
5
4
Other Zero of Polynomial
\begin{gathered}x – 5 = 0 \\ \\ x = 5\end{gathered}
x−5=0
x=5
We got Zeroes
\begin{gathered} \alpha = \frac{4}{5} \\ \\ \beta = 5\end{gathered}
α=
5
4
β=5
Polynomial
ax²+bx+c
{5x}^{2} – 29x + 205x
2
−29x+20
Coffecient of x²=5
Coffecient of x¹=-29
Coffecient of x^0=20
Let a in ax²+bx+c=5
Let B in ax²+bx+c=-29
Let c in ax²+bx+c=20
Sum Of Zeroes
\begin{gathered} \alpha + \beta = – \frac{coffecient \: of \: x}{coffecient \: of \: {x}^{2} } \\ \\ \alpha + \beta = – \frac{b}{a} \\ \\ \frac{4}{5} + 5 = – \frac{-29}{5} \\ \\ \frac{29}{4} = + \frac{29}{4} < /p > < p > \end{gathered}
α+β=−
coffecientofx
2
coffecientofx
α+β=−
a
b
5
4
+5=−
5
−29
4
29
=+
4
29
</p><p>
lhs = rhslhs=rhs
Product of Zeroes
\begin{gathered} \alpha \times \beta = \frac{constant \: term}{cofficient \: of \: {x}^{2} } \\ \\ \frac{4}{5} \times 5 = \frac{20}{5} \\ \\ \frac{20}{5} = \frac{20}{5} \\ \\ lhs = rhs\end{gathered}
α×β=
cofficientofx
2
constantterm
5
4
×5=
5
20
5
20
=
5
20
lhs=rhs
\boxed{\mathbf{\huge{LHS=RHS}}}
LHS=RHS
\underline{\mathbf{\huge{Verified✓}}}
Verified✓
Answer:
5 and 4/5
Step-by-step explanation:
5x² – 29x + 20
=》 5x² -25x -4x + 20
=》 5x(x – 5) -4(x – 5)
=》 (5x – 4) (x – 5)
=》 x=4/5 and x = 5